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Lesson 7 — Logarithmic functions

Logarithm as the inverse of the exponential. Operational properties. Natural logarithm ln and common logarithm log. Logarithmic equations. Applications: pH, Richter scale, decibel, half-life.

Used in: 1.º ano EM (15 anos) · Math I japonês cap. 4 · Klasse 10 alemã · Química (pH) · Física (decibel, Richter)

\log_a x = y \iff a^y = x

Logarithm is the inverse of the exponential: $\log_a x$ answers the question "to what power must a be raised to give x?" The base satisfies $a > 0, a \neq 1$ and the argument requires $x > 0$ .

Choose your door

Rigorous notation, full derivation, hypotheses

Definition and properties

Definition and domain

"The logarithmic function with base $b$ , $f(x) = \log_b(x)$ , is the inverse of the exponential function $g(x) = b^x$ . The domain is $(0, +\infty)$ and the range is $\mathbb{R}$ ." — OpenStax College Algebra 2e §6.3

Operational properties

Definition· Laws of logarithms

For $a > 0, a \neq 1$ and $x, y > 0$ :

\log_a (xy) = \log_a x + \log_a y

(P1)

what this means · Logarithm of a product equals the sum of the logarithms.

\log_a \!\left(\frac{x}{y}\right) = \log_a x - \log_a y

(P2)

what this means · Logarithm of a quotient equals the difference of the logarithms.

\log_a (x^n) = n\, \log_a x

(P3)

what this means · Logarithm of a power: the exponent moves in front.

\log_a x = \frac{\log_b x}{\log_b a}

(P4)

what this means · Change of base: allows computing any log using ln or log base 10.

Two fundamental identities (direct consequences of the definition):

\log_a (a^x) = x

(I1)

what this means · The logarithm undoes exponentiation: log_a of a^x equals x.

a^{\log_a x} = x

(I2)

what this means · Exponentiation undoes the logarithm: a raised to log_a of x equals x.

"The product rule for logarithms is derived directly from the property $a^m \cdot a^n = a^{m+n}$ ." — OpenStax College Algebra 2e §6.5

Graph — log and exponential as inverses

e^x and ln x are reflections of each other across the line y = x. The ln x curve passes through (1, 0) since ln 1 = 0; it grows without bound but very slowly.

Theorem and proof of the product property

Worked examples

Example— 3· Solving a logarithmic equation (intermediate)

Problem: Solve $\log_2 x + \log_2(x - 2) = 3$ .

Strategy: Combine using P1, apply the definition, and check the domain.

Solution:

Domain: we need $x > 0$ and $x - 2 > 0$ , that is $x > 2$ .
Combine: $\log_2[x(x-2)] = 3$ .
Definition: $x(x-2) = 2^3 = 8$ , that is $x^2 - 2x - 8 = 0$ .
Factor: $(x-4)(x+2) = 0 \Rightarrow x = 4$ or $x = -2$ .
Filter: $x = -2 < 2$ → discard. The solution is $x = 4$ .

Verification: $\log_2 4 + \log_2 2 = 2 + 1 = 3$ . ✓

Source. OpenStax — College Algebra 2e §6.6 — CC-BY 4.0 license.

Example— 4· Change of base with a calculator (advanced)

Problem: Compute $\log_3 17$ using only $\ln$ and $\log_{10}$ .

Strategy: Change of base formula $\log_a x = \log_b x / \log_b a$ .

Solution:

Via ln: $\log_3 17 = \dfrac{\ln 17}{\ln 3} \approx \dfrac{2{,}8332}{1{,}0986} \approx 2{,}5789$ .
Sanity check via $\log_{10}$ : $\dfrac{\log 17}{\log 3} \approx \dfrac{1{,}2304}{0{,}4771} \approx 2{,}5789$ . ✓

Verification: $3^{2{,}5789} \approx 17$ (calculator confirms). ✓

Source. Stitz–Zeager Precalculus §6.3 — CC-BY-NC-SA license.

Example— 5· pH as a logarithmic model (real-world modeling)

Problem: In a solution, $[H^+] = 4 \times 10^{-5}$ mol/L. Compute the pH. (Given: $\log_{10} 4 \approx 0{,}6$ .)

Strategy: Apply $\text{pH} = -\log_{10}[H^+]$ and use P1 to separate the factors.

Solution:

Definition: $\text{pH} = -\log_{10}(4 \times 10^{-5})$ .
P1: $\log_{10}(4 \times 10^{-5}) = \log_{10} 4 + \log_{10} 10^{-5} = 0{,}6 + (-5) = -4{,}4$ .
Sign: $\text{pH} = -(-4{,}4) = 4{,}4$ .

Interpretation: pH 4.4 — acidic solution, similar to diluted orange juice.

Verification: $10^{-4{,}4} = 10^{-5} \cdot 10^{0{,}6} \approx 10^{-5} \cdot 3{,}98 \approx 4 \times 10^{-5}$ . ✓

Source. OpenStax — Algebra and Trigonometry 2e §6.7 — CC-BY 4.0 license.

Exercise list

45 exercises · 11 with worked solution (25%)

Application 27Understanding 7Modeling 8Challenge 3

Ex. 7.1Understanding
How can the logarithmic equation $\log_b x = y$ be solved for $x$ using properties of exponents?
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The logarithmic equation $\log_b x = y$ can be solved for $x$ by rewriting it as $b^{y} = x$ .The logarithmic equation $\log_b x = y$ can be solved for $x$ by rewriting it as $x^{b} = y$ .The logarithmic equation $\log_b x = y$ can be solved for $x$ by rewriting it as $y^{b} = x$ .The logarithmic equation $\log_b x = y$ cannot be solved for $x$ without a calculator.
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 3
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The equivalent definition is $\log_b x = y \Leftrightarrow b^y = x$ . Simply raise the base to the exponent.
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1. Recognize the definition: $\log_b x = y$ means " $b$ raised to $y$ gives $x$ ".
2. Rewrite the equality in exponential form: $b^y = x$ .
3. This equivalence is exactly the definition of logarithm as the inverse of the exponential.
Ex. 7.2Understanding
Discuss the meaning of the common logarithm. What is its relationship to a base- $b$ logarithm and how does the notation differ?
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The common logarithm is $\log_{10} x$ , written simply as $\log x$ .The common logarithm is $\log_e x = \ln x$ , base $e$ .The common logarithm has any base $b$ as long as $b > 0$ .The common logarithm uses the notation $\ln x$ and has base $e$ .
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 4
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Common logarithm = base 10. The notation is $\log x$ (no base subscript), in contrast to $\ln x$ which has base $e$ .
Ex. 7.3UnderstandingAnswer key
What types of translations affect the domain of a logarithmic function?
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Horizontal translations can change the domain, but not the range.Horizontal translations change both the domain and the range.Horizontal translations change neither the domain nor the range.Horizontal translations change the range, but not the domain.
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Solve online OpenStax College Algebra 2e · §6.4 · ex. 3
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Shifting the curve horizontally (replacing $x$ with $x - h$ ) moves the vertical asymptote and changes the domain, but the range remains $\mathbb{R}$ .
Ex. 7.4Understanding
Consider the general logarithmic function $f(x) = \log_b(x)$ . Why can $x$ not be zero?
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$x > 0$ , since the argument of the logarithm must be positive. $x \geq 0$ , since the logarithm of zero is zero. $x \neq 0$ , since the logarithm of zero is undefined. $x$ can be any real number, since the function is continuous.
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OpenStax College Algebra 2e · §6.4 · ex. 4
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For $f(x) = \log_b(x)$ to be defined, the argument $x$ must be strictly positive: $x > 0$ . The logarithm of zero and of negative numbers does not exist in the reals.
Ex. 7.5Application
Rewrite the equation $\log_4(q) = m$ in exponential form.
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$4^m = q$ $q^4 = m$ $4^q = m$ $m^4 = q$
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 6
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Exponential form of $\log_4(q) = m$ : base 4, exponent $m$ , result $q$ . Therefore $4^m = q$ .
Ex. 7.6Application
Rewrite $\log_a(b) = c$ in exponential form.
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$a^c = b$ $b^a = c$ $c^a = b$ $a^b = c$
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 7
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Exponential form of $\log_a(b) = c$ : $a^c = b$ .
Ex. 7.7Application
Rewrite $\log_{16}(y) = x$ in exponential form.
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$16^x = y$ $y^{16} = x$ $x^{16} = y$ $16^y = x$
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 8
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Exponential form of $\log_{16}(y) = x$ : $16^x = y$ .
Ex. 7.8Application
Rewrite $\log_x(64) = y$ in exponential form.
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$x^y = 64$ $64^x = y$ $y^x = 64$ $x^{64} = y$
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 9
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Exponential form of $\log_x(64) = y$ : base $x$ , exponent $y$ , result $64$ . Therefore $x^y = 64$ .
Ex. 7.9Application
Rewrite $\log_y(x) = -11$ in exponential form.
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$y^{-11} = x$ $x^{-11} = y$ $y^x = -11$ $x^y = -11$
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 10
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Exponential form of $\log_y(x) = -11$ : $y^{-11} = x$ .
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1. Identify: base $y$ , exponent $-11$ , result $x$ .
2. Write in the form $\text{base}^{\text{exponent}} = \text{result}$ .
3. Conclusion: $y^{-11} = x$ .
Ex. 7.10Application
Rewrite $2^x = y$ in logarithmic form.
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$\log_2 y = x$ $\log_y 2 = x$ $\log_2 x = y$ $\log_x y = 2$
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 14
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Logarithmic form of $2^x = y$ : base 2, exponent $x$ , result $y$ . Therefore $\log_2 y = x$ .
Ex. 7.11Application
Rewrite $10^a = b$ in logarithmic form.
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$\log_{10} b = a$ $\log_b 10 = a$ $\log_{10} a = b$ $\log_a b = 10$
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 22
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Logarithmic form of $10^a = b$ : $\log_{10} b = a$ .
Ex. 7.12Application
Evaluate $\log_5 25$ without a calculator.
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$2$ $\frac{1}{2}$ $-2$ $4$
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 30
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Since $5^2 = 25$ , we have $\log_5 25 = 2$ . (Ans: 2)
Ex. 7.13Application
Evaluate $\log_2 8$ without a calculator.
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$3$ $4$ $-3$ $9$
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 29
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Since $2^3 = 8$ , we have $\log_2 8 = 3$ . (Ans: 3)
Ex. 7.14ApplicationAnswer key
Evaluate $\log(0{,}0001)$ without a calculator.
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$-4$ $4$ $\frac{1}{4}$ $-\frac{1}{4}$
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 34
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Since $10^{-4} = 0{,}0001$ , we have $\log(0{,}0001) = -4$ . (Ans: -4)
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1. Write $0{,}0001 = 10^{-4}$ .
2. Apply the identity: $\log_{10}(10^{-4}) = -4$ .
3. Conclusion: $\log(0{,}0001) = -4$ .
Ex. 7.15Application
Evaluate $\ln(e^{-3})$ without a calculator.
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$-3$ $3$ $\frac{1}{3}$ $-\frac{1}{3}$
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 46
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Since $e^{-3}$ is the exponent: $\ln(e^{-3}) = -3$ by the inverse identity. (Ans: -3)
Ex. 7.16Application
Evaluate $\ln(e^{1/3})$ without a calculator.
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$\frac{1}{3}$ $3$ $-3$ $\frac{1}{9}$
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 50
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By the identity: $\ln(e^{1/3}) = \frac{1}{3}$ . (Ans: 1/3)
Ex. 7.17ApplicationAnswer key
Evaluate $\ln(1)$ without a calculator.
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$0$ $1$ $e$ undefined
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 51
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By definition: $a^0 = 1$ for any $a > 0$ , therefore $\ln(1) = 0$ . (Ans: 0)
Ex. 7.18ApplicationAnswer key
Evaluate $\ln(e^{-0{,}225}) - 3$ without a calculator.
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$-3{,}225$ $-2{,}225$ $3{,}225$ $0{,}225$
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 52
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By the identity: $\ln(e^{-0{,}225}) = -0{,}225$ . Subtracting 3: $-0{,}225 - 3 = -3{,}225$ . (Ans: -3.225)
Ex. 7.19Application
Evaluate $25\ln(e^{2/5})$ without a calculator.
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$10$ $\frac{2}{5}$ $5$ $25$
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 53
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By the identity $\ln(e^x) = x$ : $25 \ln(e^{2/5}) = 25 \cdot \frac{2}{5} = 10$ . (Ans: 10)
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1. Use the identity: $\ln(e^{2/5}) = \frac{2}{5}$ .
2. Multiply by the coefficient: $25 \cdot \frac{2}{5} = 10$ .
Ex. 7.20ModelingAnswer key
Determine the domain, range, and vertical asymptote of $h(x) = \log_4(x-1) + 1$ .
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Domain $(1,\infty)$ , range $\mathbb{R}$ , vertical asymptote $x = 1$ Domain $(-\infty,1)$ , range $\mathbb{R}$ , vertical asymptote $x = 1$ Domain $(1,\infty)$ , range $[0,\infty)$ , vertical asymptote $x = 1$ Domain $[1,\infty)$ , range $\mathbb{R}$ , vertical asymptote $x = 0$
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Solve online OpenStax College Algebra 2e · §6.4 · ex. 21
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For $h(x) = \log_4(x-1) + 1$ : a positive argument requires $x - 1 > 0$ , so domain is $(1, \infty)$ ; the range of any logarithm is $\mathbb{R}$ ; vertical asymptote at $x = 1$ .
Ex. 7.21Modeling
Determine the domain and vertical asymptote of $f(x) = \log(5x+10) + 3$ .
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Domain $(-2,\infty)$ , asymptote $x = -2$ Domain $(-\infty,-2)$ , asymptote $x = -2$ Domain $(-2,\infty)$ , asymptote $x = 0$ Domain $[-2,\infty)$ , asymptote $x = -2$
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Solve online OpenStax College Algebra 2e · §6.4 · ex. 22
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For $f(x) = \log(5x+10) + 3$ : a positive argument requires $5x + 10 > 0 \Rightarrow x > -2$ ; vertical asymptote at $x = -2$ .
Ex. 7.22Modeling
Determine the domain and vertical asymptote of $g(x) = \ln(-x) - 2$ .
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Domain $(-\infty,0)$ , asymptote $x = 0$ Domain $(0,\infty)$ , asymptote $x = 0$ Domain $(-\infty,0)$ , asymptote $x = -2$ Domain $\mathbb{R}$ , no vertical asymptote
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Solve online OpenStax College Algebra 2e · §6.4 · ex. 23
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For $g(x) = \ln(-x) - 2$ : a positive argument requires $-x > 0 \Rightarrow x < 0$ ; vertical asymptote at $x = 0$ .
Ex. 7.23Modeling
Determine the domain and vertical asymptote of $f(x) = \log(3x+1)$ .
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Domain $(-\frac{1}{3},\infty)$ , asymptote $x = -\frac{1}{3}$ Domain $(\frac{1}{3},\infty)$ , asymptote $x = \frac{1}{3}$ Domain $(-\frac{1}{3},\infty)$ , asymptote $x = 0$ Domain $(-\infty,-\frac{1}{3})$ , asymptote $x = -\frac{1}{3}$
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Solve online OpenStax College Algebra 2e · §6.4 · ex. 13
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For $f(x) = \log(3x+1)$ : a positive argument requires $3x + 1 > 0 \Rightarrow x > -\frac{1}{3}$ ; vertical asymptote at $x = -\frac{1}{3}$ .
Ex. 7.24ModelingAnswer key
Determine the domain and vertical asymptote of $g(x) = -\ln(3x+9) - 7$ .
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Domain $(-3,\infty)$ , asymptote $x = -3$ Domain $(-\infty,-3)$ , asymptote $x = -3$ Domain $(-3,\infty)$ , asymptote $x = 0$ Domain $[-3,\infty)$ , asymptote $x = -3$
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Solve online OpenStax College Algebra 2e · §6.4 · ex. 15
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For $g(x) = -\ln(3x+9) - 7$ : a positive argument requires $3x + 9 > 0 \Rightarrow x > -3$ ; vertical asymptote at $x = -3$ .
Ex. 7.25Understanding
Let $b$ be any positive real number such that $b \neq 1$ . What must $\log_b 1$ equal? Justify your answer.
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$\log_b 1 = 0$ for any valid base $b > 0, b \neq 1$ . $\log_b 1 = 1$ for any valid base. $\log_b 1 = b$ for any valid base. $\log_b 1$ is undefined since 1 is not positive.
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Solve online OpenStax College Algebra 2e · §6.4 · ex. 56
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For any valid base, $b^0 = 1$ , therefore $\log_b 1 = 0$ . This follows directly from the definition of logarithm as the inverse of the exponential.
Show step-by-step (with the why)
1. Let $b > 0, b \neq 1$ be any valid base.
2. Find the exponent $y$ such that $b^y = 1$ .
3. By the property of exponents, $b^0 = 1$ for all $b$ .
4. Therefore $\log_b 1 = 0$ .
Ex. 7.26Understanding
How does the power rule for logarithms help when computing $\log_b(x^n)$ ?
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The power rule (P3) brings the exponent forward as a multiplicative factor.The power rule divides the exponent by the logarithm of the base.The power rule converts the logarithm of a power into a product of logarithms.The power rule only works when the exponent is an integer.
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Solve online OpenStax College Algebra 2e · §6.5 · ex. 1
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The power rule: $\log_b(x^n) = n \log_b x$ . It brings the exponent $n$ forward as a factor, simplifying computations with arbitrary powers.
Ex. 7.27Understanding
What does the change of base formula do? Why is it useful when using a calculator?
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Allows computing logarithms in any base using only $\ln$ or $\log_{10}$ .Allows simplifying the product of two logarithms into a single logarithm.Allows computing logarithms of products without using the product rule.Always converts logarithms to base $e$ , regardless of the original base.
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Solve online OpenStax College Algebra 2e · §6.5 · ex. 2
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The change of base formula: $\log_a x = \frac{\ln x}{\ln a} = \frac{\log x}{\log a}$ . It is useful because scientific calculators only have $\ln$ and $\log_{10}$ .
Ex. 7.28Application
Expand as completely as possible: $\log_b(7x \cdot 2y)$ .
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$\log_b 7 + \log_b x + \log_b 2 + \log_b y$ $\log_b(7x) - \log_b(2y)$ $7 \log_b x + 2 \log_b y$ $\log_b(7x \cdot 2y)$ cannot be expanded
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Solve online OpenStax College Algebra 2e · §6.5 · ex. 3
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By P1: $\log_b(7x \cdot 2y) = \log_b 7 + \log_b x + \log_b 2 + \log_b y$ .
Ex. 7.29ApplicationAnswer key
Expand as completely as possible: $\ln(3ab \cdot 5c)$ .
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$\ln 3 + \ln a + \ln b + \ln 5 + \ln c$ $\ln 3 + \ln a + \ln b - \ln 5 - \ln c$ $\ln(3ab) + \ln(5c)$ $3\ln a + 5\ln c$
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Solve online OpenStax College Algebra 2e · §6.5 · ex. 4
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By P1 applied repeatedly: $\ln(3ab \cdot 5c) = \ln 3 + \ln a + \ln b + \ln 5 + \ln c$ .
Ex. 7.30ApplicationAnswer key
Expand as completely as possible: $\log_b\!\left(\frac{13}{17}\right)$ .
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$\log_b 13 - \log_b 17$ $\log_b 13 + \log_b 17$ $\frac{\log_b 13}{\log_b 17}$ $\frac{13}{17}\log_b 1$
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Solve online OpenStax College Algebra 2e · §6.5 · ex. 5
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By P2: $\log_b\!\left(\frac{13}{17}\right) = \log_b 13 - \log_b 17$ .
Ex. 7.31Application
Expand as completely as possible: $\ln\!\left(\frac{1}{4k}\right)$ .
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$-\ln 4 - \ln k$ $\ln 4 + \ln k$ $\ln 4 - \ln k$ $-\frac{\ln k}{\ln 4}$
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Solve online OpenStax College Algebra 2e · §6.5 · ex. 7
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By P2: $\ln\!\left(\frac{1}{4k}\right) = \ln 1 - \ln 4 - \ln k = 0 - \ln 4 - \ln k = -\ln 4 - \ln k$ .
Show step-by-step (with the why)
1. Write as a quotient: $\ln(1) - \ln(4k)$ .
2. Since $\ln 1 = 0$ : result is $-\ln(4k)$ .
3. Apply P1: $-\ln 4 - \ln k$ .
Ex. 7.32Application
Condense into a single logarithm: $\ln 7 + \ln x + \ln y$ .
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$\ln(7xy)$ $7 \ln(xy)$ $7 + \ln(xy)$ $\ln(7 + x + y)$
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Solve online OpenStax College Algebra 2e · §6.5 · ex. 9
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By P1: $\ln 7 + \ln x + \ln y = \ln(7xy)$ .
Ex. 7.33Application
Condense into a single logarithm: $\log_b 28 - \log_b 7$ .
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$\log_b 4$ $\log_b(28 + 7)$ $4\log_b 7$ $\log_b\!\left(\frac{7}{28}\right)$
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Solve online OpenStax College Algebra 2e · §6.5 · ex. 11
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By P2: $\log_b 28 - \log_b 7 = \log_b\!\left(\frac{28}{7}\right) = \log_b 4$ .
Ex. 7.34Application
Condense into a single logarithm: $\frac{1}{3}\ln 8$ .
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$\ln 2$ $\ln 8$ $3\ln 2$ $\frac{1}{3}\ln 8$
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Solve online OpenStax College Algebra 2e · §6.5 · ex. 14
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By P3: $\frac{1}{3}\ln 8 = \ln(8^{1/3}) = \ln 2$ . (Ans: $\ln 2$ )
Ex. 7.35ApplicationAnswer key
Evaluate without a calculator: $\log_3\!\left(\frac{1}{9}\right) - 3\log_3 3$ .
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$-5$ $5$ $-2$ $2$
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Solve online OpenStax College Algebra 2e · §6.5 · ex. 30
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By P3: $\log_3\!\left(\frac{1}{9}\right) = \log_3(3^{-2}) = -2$ . Subtracting: $-2 - 3\log_3 3 = -2 - 3 = -5$ . (Ans: -5)
Show step-by-step (with the why)
1. $\log_3(1/9) = \log_3(3^{-2}) = -2$ .
2. $\log_3 3 = 1$ , so $3\log_3 3 = 3$ .
3. Result: $-2 - 3 = -5$ .
Ex. 7.36Application
Evaluate without a calculator: $2\log_9 3 - 4\log_9 3 + \log_9\!\left(\frac{1}{729}\right)$ . (Ans: -4)
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$-4$ $-2$ $0$ $4$
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Solve online OpenStax College Algebra 2e · §6.5 · ex. 32
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Combine the two terms with $\log_9 3$ : $(2 - 4)\log_9 3 = -2 \cdot \frac{1}{2} = -1$ . Then, $\log_9\!\left(\frac{1}{729}\right) = \log_9(3^{-6}) = -6 \cdot \frac{1}{2} = -3$ . Total: $-1 + (-3) = -4$ . (Ans: -4)
Ex. 7.37Application
Use the change of base formula to evaluate $\log_3 22$ as a quotient of natural logarithms. Round to five decimal places.
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$\approx 2{,}81359$ $\approx 1{,}40680$ $\approx 3{,}09104$ $\approx 0{,}72792$
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Solve online OpenStax College Algebra 2e · §6.5 · ex. 33
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Change of base: $\log_3 22 = \frac{\ln 22}{\ln 3} \approx \frac{3{,}09104}{1{,}09861} \approx 2{,}81359$ . (Ans: approx. 2.81359)
Ex. 7.38Application
Use the change of base formula to evaluate $\log_8 65$ as a quotient of natural logarithms. Round to five decimal places.
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$\approx 2{,}00746$ $\approx 4{,}17439$ $\approx 0{,}49814$ $\approx 1{,}81291$
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Solve online OpenStax College Algebra 2e · §6.5 · ex. 34
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Change of base: $\log_8 65 = \frac{\ln 65}{\ln 8} \approx \frac{4{,}17439}{2{,}07944} \approx 2{,}00746$ . (Ans: approx. 2.00746)
Ex. 7.39ApplicationAnswer key
Use the change of base formula to evaluate $\log_6 5{,}38$ as a quotient of natural logarithms. Round to five decimal places.
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$\approx 0{,}93913$ $\approx 1{,}08257$ $\approx 1{,}68245$ $\approx 0{,}61939$
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Solve online OpenStax College Algebra 2e · §6.5 · ex. 35
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Change of base: $\log_6 5{,}38 = \frac{\ln 5{,}38}{\ln 6} \approx \frac{1{,}68245}{1{,}79176} \approx 0{,}93913$ . (Ans: approx. 0.93913)
Ex. 7.40Challenge
Use the product rule for logarithms to find all values of $x$ such that $\log_{12}(2x+6) + \log_{12}(x+2) = 2$ . Show your solution steps.
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$x = 6$ (the only value satisfying the equation and the domain) $x = -11$ and $x = 6$ $x = -11$ onlyNo real value satisfies the equation
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Solve online OpenStax College Algebra 2e · §6.5 · ex. 38
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By P1: $(2x+6)(x+2) = 144$ . Expanding: $2x^2 + 10x - 132 = 0 \Rightarrow x^2 + 5x - 66 = 0$ . Discriminant: $25 + 264 = 289 = 17^2$ . Roots: $x = \frac{-5 \pm 17}{2}$ , so $x = 6$ or $x = -11$ . The domain requires $x > -2$ , therefore $x = 6$ is the only solution.
Show step-by-step (with the why)
1. Apply P1: $\log_{12}[(2x+6)(x+2)] = 2$ .
2. Convert: $(2x+6)(x+2) = 12^2 = 144$ .
3. Expand: $2x^2 + 10x + 12 = 144 \Rightarrow x^2 + 5x - 66 = 0$ .
4. Quadratic formula: $x = \frac{-5 \pm \sqrt{289}}{2} = \frac{-5 \pm 17}{2}$ , so $x = 6$ or $x = -11$ .
5. Domain: $2x + 6 > 0 \Rightarrow x > -3$ and $x + 2 > 0 \Rightarrow x > -2$ . Accept $x = 6$ .
Ex. 7.41Challenge
Use the quotient rule for logarithms to find all values of $x$ such that $\log_6(x+2) - \log_6(x-3) = 1$ . (Ans: $x = 4$ )
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$x = 4$ $x = 5$ $x = -5$ $x = \frac{5}{17}$
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Solve online OpenStax College Algebra 2e · §6.5 · ex. 39
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By P2: $\log_6\!\left(\frac{x+2}{x-3}\right) = 1 \Rightarrow \frac{x+2}{x-3} = 6$ . Solving: $x + 2 = 6x - 18 \Rightarrow 20 = 5x \Rightarrow x = 4$ . Checking the domain with $x = 4$ : $x + 2 = 6 > 0$ and $x - 3 = 1 > 0$ ; valid. (Ans: 4)
Show step-by-step (with the why)
1. Apply P2: $\log_6\!\left(\frac{x+2}{x-3}\right) = 1$ .
2. Convert: $\frac{x+2}{x-3} = 6^1 = 6$ .
3. Solve: $x + 2 = 6(x-3) = 6x - 18 \Rightarrow 20 = 5x \Rightarrow x = 4$ .
4. Check domain: both arguments positive at $x = 4$ . Valid.
Ex. 7.42Modeling
The exposure index $EI$ of a camera is given by $EI = \log_2(f^2 t)$ , where $f$ is the aperture (f-stop) and $t$ is the exposure time in seconds. If $f = 8$ and $t = 2$ , what is the exposure index? (Ans: 7)
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7685
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 64
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With $f = 8$ and $t = 2$ : $EI = \log_2(f^2 t) = \log_2(64 \cdot 2) = \log_2 128 = \log_2(2^7) = 7$ .
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1. Substitute: $EI = \log_2(8^2 \cdot 2) = \log_2(64 \cdot 2) = \log_2 128$ .
2. Write 128 as a power of 2: $128 = 2^7$ .
3. Apply the identity: $\log_2(2^7) = 7$ .
Ex. 7.43Modeling
Using the formula $EI = \log_2(f^2 t)$ : if the meter reads $EI = -2$ and the desired exposure time is $t = 16$ s, what should the aperture $f$ be? (Ans: $f = 1/8$ )
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$f = \frac{1}{8}$ $f = 8$ $f = \frac{1}{4}$ $f = 2$
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 65
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With $EI = -2$ and $t = 16$ : $-2 = \log_2(f^2 \cdot 16) \Rightarrow f^2 \cdot 16 = 2^{-2} = \frac{1}{4} \Rightarrow f^2 = \frac{1}{64} \Rightarrow f = \frac{1}{8}$ .
Ex. 7.44Modeling
The intensities of two earthquakes at a seismographic station can be compared using the formula $\log\!\left(\frac{I_1}{I_2}\right) = M_1 - M_2$ . In August 2009 an earthquake of magnitude 6.1 occurred; in March 2011, one of magnitude 9.0. How many times more intense was the second compared to the first? (Ans: approx. 794)
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Approximately $794$ times more intenseApproximately $2{,}9$ times more intenseApproximately $0{,}0013$ times more intenseApproximately $100$ times more intense
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Solve online OpenStax College Algebra 2e · §6.3 · ex. 66
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Using $\log\!\left(\frac{I_1}{I_2}\right) = M_1 - M_2$ with $M_1 = 6{,}1$ and $M_2 = 9{,}0$ : $\frac{I_2}{I_1} = 10^{M_2 - M_1} = 10^{2{,}9} \approx 794$ . The magnitude-9.0 earthquake was about 794 times more intense.
Show step-by-step (with the why)
1. Rewrite: $\log(I_2/I_1) = M_2 - M_1 = 9{,}0 - 6{,}1 = 2{,}9$ .
2. Convert: $I_2/I_1 = 10^{2{,}9} \approx 794$ .
3. Interpretation: the second earthquake was approximately 794 times more intense.
Ex. 7.45ChallengeAnswer key
Prove that $\log_b(n) = \frac{1}{\log_n(b)}$ for any positive integers $b > 1$ and $n > 1$ .
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$\log_b(n) = \frac{1}{\log_n(b)}$ for integers $b > 1$ and $n > 1$ . $\log_b(n) = \log_n(b)$ for any positive bases. $\log_b(n) = -\log_n(b)$ since the relation is inverse. $\log_b(n) \cdot \log_n(b) = 0$ by the product property.
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OpenStax College Algebra 2e · §6.5 · ex. 41
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By the change of base formula: $\log_b(n) = \frac{\ln n}{\ln b}$ and $\log_n(b) = \frac{\ln b}{\ln n}$ . Therefore $\log_b(n) = \frac{1}{\log_n(b)}$ — they are reciprocals of each other.

Sources

Only books that directly fed this text and exercises. Full catalog at /livros.

OpenStax College Algebra 2e — Jay Abramson et al. · 2022, 2nd ed · EN · CC-BY 4.0 · §6.3–6.6 (definition, operational properties, logarithmic equations). Primary source for blocks A, B, and C.
OpenStax Algebra and Trigonometry 2e — Jay Abramson et al. · 2022, 2nd ed · EN · CC-BY 4.0 · §6.3–6.7 (logarithms, properties, exponential and logarithmic models). Primary source for block D.
Stitz–Zeager Precalculus — Carl Stitz, Jeff Zeager · 2013, v3 · EN · CC-BY-NC-SA · §6.2–6.3 (logarithmic functions, change of base, equations). Source for block E (change of base).
Active Calculus 2.0 — Matt Boelkins · 2024 · EN · CC-BY-NC-SA · §1.7 (natural logarithm and its relation to $1/x$ ). Exercise 7.46.