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Step-by-Step Guidance for Logarithmic & Exponential Equations (Precalculus)

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Q1. Solve for $x$: $\log_7(9x - 4) = \log_7(20 + x)$

Background

Topic: Logarithmic Equations

This question tests your ability to solve equations involving logarithms with the same base. You will use properties of logarithms to simplify and solve for the variable.

Key Terms and Formulas:

  • Logarithm: $\log_b(a)$ is the exponent to which $b$ must be raised to get $a$.

  • One-to-One Property: If $\log_b(A) = \log_b(B)$, then $A = B$ (as long as $A, B > 0$).

Step-by-Step Guidance

  1. Since both sides have $\log_7$, set the arguments equal: $9x - 4 = 20 + x$.

  2. Isolate $x$ by subtracting $x$ from both sides: $9x - x - 4 = 20$.

  3. Simplify the equation to get $8x - 4 = 20$.

  4. Add $4$ to both sides to isolate the term with $x$.

Try solving on your own before revealing the answer!

Final Answer: $x = 3$

Add $4$ to both sides: $8x = 24$, then divide by $8$ to get $x = 3$.

Check: $\log_7(9 \times 3 - 4) = \log_7(20 + 3) \implies \log_7(23) = \log_7(23)$, which is valid.

Q2. Solve for $m$: $\log_5(2m - 5) = \log_5(12)$

Background

Topic: Logarithmic Equations

This question tests your understanding of solving logarithmic equations by equating arguments when the logs have the same base.

Key Terms and Formulas:

  • One-to-One Property: If $\log_b(A) = \log_b(B)$, then $A = B$ (for $A, B > 0$).

Step-by-Step Guidance

  1. Set the arguments equal: $2m - 5 = 12$.

  2. Add $5$ to both sides to isolate the term with $m$.

  3. Solve for $m$ by dividing both sides by $2$.

Try solving on your own before revealing the answer!

Final Answer: $m = 8.5$

After adding $5$ and dividing by $2$, $m = 8.5$. Check: $2 \times 8.5 - 5 = 12$.

Q3. Solve for $a$: $\log_3(4a) + \log_3(5) = \log_3(56)$

Background

Topic: Logarithmic Equations (Properties of Logarithms)

This question tests your ability to use properties of logarithms to combine and solve equations.

Key Terms and Formulas:

  • Product Property: $\log_b(A) + \log_b(B) = \log_b(AB)$

  • One-to-One Property: If $\log_b(A) = \log_b(B)$, then $A = B$.

Step-by-Step Guidance

  1. Combine the left side using the product property: $\log_3(4a \times 5) = \log_3(56)$.

  2. Simplify the argument: $\log_3(20a) = \log_3(56)$.

  3. Set the arguments equal: $20a = 56$.

  4. Divide both sides by $20$ to solve for $a$.

Try solving on your own before revealing the answer!

Final Answer: $a = 2.8$

Dividing both sides by $20$ gives $a = 2.8$. Check: $4 \times 2.8 = 11.2$, $\log_3(11.2) + \log_3(5) = \log_3(56)$.

Q4. Solve for $y$: $\log_2(10y) - 7\log_2(y) = \log_2(8)$

Background

Topic: Logarithmic Equations (Properties of Logarithms)

This question tests your ability to use properties of logarithms, including the power and quotient rules, to solve for a variable.

Key Terms and Formulas:

  • Power Rule: $k\log_b(A) = \log_b(A^k)$

  • Quotient Rule: $\log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right)$

Step-by-Step Guidance

  1. Rewrite $7\log_2(y)$ as $\log_2(y^7)$ using the power rule.

  2. Apply the quotient rule: $\log_2\left(\frac{10y}{y^7}\right) = \log_2(8)$.

  3. Simplify the argument: $\frac{10y}{y^7} = 10y^{1-7} = 10y^{-6}$.

  4. Set the arguments equal: $10y^{-6} = 8$.

Try solving on your own before revealing the answer!

Final Answer: $y = \sqrt[6]{\frac{10}{8}}$

Set $10y^{-6} = 8$, so $y^{-6} = \frac{8}{10}$, then solve for $y$ by taking the reciprocal and the $6$th root.

Q5. Solve for $m$: $\log_4(5m + 9) = 3$

Background

Topic: Logarithmic Equations (Converting to Exponential Form)

This question tests your ability to rewrite a logarithmic equation in exponential form to solve for the variable.

Key Terms and Formulas:

  • Definition of Logarithm: $\log_b(A) = C \iff b^C = A$

Step-by-Step Guidance

  1. Rewrite the equation in exponential form: $4^3 = 5m + 9$.

  2. Calculate $4^3$.

  3. Subtract $9$ from both sides to isolate $5m$.

  4. Divide both sides by $5$ to solve for $m$.

Try solving on your own before revealing the answer!

Final Answer: $m = 11$

$4^3 = 64$, so $5m = 64 - 9 = 55$, $m = 11$.

Q6. Solve for $p$: $\log_{36}(20p - 4) = \frac{1}{2}$

Background

Topic: Logarithmic Equations (Converting to Exponential Form)

This question tests your ability to convert a logarithmic equation to exponential form and solve for the variable.

Key Terms and Formulas:

  • Definition of Logarithm: $\log_b(A) = C \iff b^C = A$

Step-by-Step Guidance

  1. Rewrite the equation in exponential form: $36^{1/2} = 20p - 4$.

  2. Recall that $36^{1/2}$ means the square root of $36$.

  3. Add $4$ to both sides to isolate $20p$.

  4. Divide both sides by $20$ to solve for $p$.

Try solving on your own before revealing the answer!

Final Answer: $p = 0.5$

$36^{1/2} = 6$, so $20p = 6 + 4 = 10$, $p = 0.5$.

Q7. Solve for $k$: $\log_6(7k - 1) = 3$

Background

Topic: Logarithmic Equations (Converting to Exponential Form)

This question tests your ability to convert a logarithmic equation to exponential form and solve for the variable.

Key Terms and Formulas:

  • Definition of Logarithm: $\log_b(A) = C \iff b^C = A$

Step-by-Step Guidance

  1. Rewrite the equation in exponential form: $6^3 = 7k - 1$.

  2. Calculate $6^3$.

  3. Add $1$ to both sides to isolate $7k$.

  4. Divide both sides by $7$ to solve for $k$.

Try solving on your own before revealing the answer!

Final Answer: $k = 37$

$6^3 = 216$, so $7k = 216 + 1 = 217$, $k = 31$.

Q8. Solve for $n$: $\log_8(n + 2) + \log_4(2) = 0$

Background

Topic: Logarithmic Equations (Change of Base and Properties)

This question tests your ability to use properties of logarithms and change of base to solve for a variable.

Key Terms and Formulas:

  • Change of Base Formula: $\log_b(a) = \frac{\log_k(a)}{\log_k(b)}$

  • Product Property: $\log_b(A) + \log_b(B) = \log_b(AB)$ (if same base)

Step-by-Step Guidance

  1. Rewrite $\log_4(2)$ using the change of base formula: $\log_4(2) = \frac{1}{2}$.

  2. Substitute into the equation: $\log_8(n + 2) + \frac{1}{2} = 0$.

  3. Subtract $\frac{1}{2}$ from both sides: $\log_8(n + 2) = -\frac{1}{2}$.

  4. Rewrite in exponential form: $n + 2 = 8^{-1/2}$.

Try solving on your own before revealing the answer!

Final Answer: $n = \frac{1}{\sqrt{8}} - 2$

$8^{-1/2} = \frac{1}{\sqrt{8}}$, so $n = \frac{1}{\sqrt{8}} - 2$.

Q9. Solve for $v$: $2^{v-2} = 625$

Background

Topic: Exponential Equations

This question tests your ability to solve exponential equations by expressing both sides with the same base or using logarithms.

Key Terms and Formulas:

  • Exponential Equation: $a^{x} = b$

  • Logarithm: $x = \log_a(b)$

Step-by-Step Guidance

  1. Express $625$ as a power of $2$ if possible, or take the logarithm of both sides.

  2. Take $\log_2$ of both sides: $v - 2 = \log_2(625)$.

  3. Solve for $v$ by adding $2$ to both sides.

Try solving on your own before revealing the answer!

Final Answer: $v = \log_2(625) + 2$

Use a calculator to evaluate $\log_2(625)$, then add $2$.

Q10. Solve for $x$: $4^{2x-1} = 8^{x-2}$

Background

Topic: Exponential Equations (Same Base)

This question tests your ability to rewrite both sides of an equation with the same base and solve for the variable.

Key Terms and Formulas:

  • Exponential Properties: $a^{m} = (a^{n})^{m/n}$

  • Change of Base: $8 = 2^3$, $4 = 2^2$

Step-by-Step Guidance

  1. Rewrite $4$ and $8$ as powers of $2$: $4 = 2^2$, $8 = 2^3$.

  2. Substitute: $(2^2)^{2x-1} = (2^3)^{x-2}$.

  3. Simplify exponents: $2^{4x-2} = 2^{3x-6}$.

  4. Set exponents equal: $4x - 2 = 3x - 6$.

Try solving on your own before revealing the answer!

Final Answer: $x = -4$

Subtract $3x$ from both sides: $x - 2 = -6$, so $x = -4$.

Q11. Solve for $k$: $8^k = 78$

Background

Topic: Exponential Equations (Logarithms)

This question tests your ability to solve for the exponent using logarithms.

Key Terms and Formulas:

  • Logarithm: $k = \log_8(78)$

Step-by-Step Guidance

  1. Take $\log_8$ of both sides: $k = \log_8(78)$.

  2. Alternatively, use the change of base formula: $k = \frac{\log(78)}{\log(8)}$.

Try solving on your own before revealing the answer!

Final Answer: $k \approx 1.972$

Use a calculator to evaluate $k = \frac{\log(78)}{\log(8)}$.

Q12. Solve for $m$: $6^{9-m} = 78$

Background

Topic: Exponential Equations (Logarithms)

This question tests your ability to solve for the exponent using logarithms.

Key Terms and Formulas:

  • Logarithm: $x = \log_b(a)$

Step-by-Step Guidance

  1. Take $\log_6$ of both sides: $9 - m = \log_6(78)$.

  2. Solve for $m$ by subtracting $\log_6(78)$ from $9$.

  3. Isolate $m$ by multiplying both sides by $-1$ if necessary.

Try solving on your own before revealing the answer!

Final Answer: $m = 9 - \log_6(78)$

Use a calculator to evaluate $\log_6(78)$ and subtract from $9$.

Q13. Solve for $a$: $3^{15 + 7a} = 67$

Background

Topic: Exponential Equations (Logarithms)

This question tests your ability to solve for the exponent using logarithms.

Key Terms and Formulas:

  • Logarithm: $x = \log_b(a)$

Step-by-Step Guidance

  1. Take $\log_3$ of both sides: $15 + 7a = \log_3(67)$.

  2. Subtract $15$ from both sides to isolate $7a$.

  3. Divide both sides by $7$ to solve for $a$.

Try solving on your own before revealing the answer!

Final Answer: $a = \frac{\log_3(67) - 15}{7}$

Use a calculator to evaluate $\log_3(67)$, subtract $15$, then divide by $7$.

Q14. Solve for $x$: $3^{8 - 14x} + 9^{7x} = 77$

Background

Topic: Exponential Equations (Combining Like Bases)

This question tests your ability to rewrite exponents with the same base and solve for the variable.

Key Terms and Formulas:

  • Exponential Properties: $9 = 3^2$

Step-by-Step Guidance

  1. Rewrite $9^{7x}$ as $(3^2)^{7x} = 3^{14x}$.

  2. Now the equation is $3^{8 - 14x} + 3^{14x} = 77$.

  3. Let $y = 3^{14x}$, so the equation becomes $3^8 \cdot y^{-1} + y = 77$.

  4. Multiply both sides by $y$ to clear the denominator and solve the resulting quadratic equation in $y$.

Try solving on your own before revealing the answer!

Final Answer: $x = \frac{1}{14} \log_3(y)$, where $y$ is the solution to the quadratic equation $y^2 - 77y + 6561 = 0$

Solve the quadratic for $y$, then back-substitute to find $x$.

Q15. Solve for $n$: $8^{3-n} \cdot 21^{5-n} = 51$

Background

Topic: Exponential Equations (Logarithms)

This question tests your ability to use logarithms to solve for a variable in an equation with multiple exponential terms.

Key Terms and Formulas:

  • Logarithm: $\log(a \cdot b) = \log(a) + \log(b)$

Step-by-Step Guidance

  1. Take the natural logarithm (or log base 10) of both sides: $\ln(8^{3-n} \cdot 21^{5-n}) = \ln(51)$.

  2. Use properties of logarithms: $\ln(8^{3-n}) + \ln(21^{5-n}) = \ln(51)$.

  3. Bring exponents down: $(3-n)\ln(8) + (5-n)\ln(21) = \ln(51)$.

  4. Expand and collect like terms in $n$.

Try solving on your own before revealing the answer!

Final Answer: $n = \frac{3\ln(8) + 5\ln(21) - \ln(51)}{\ln(8) + \ln(21)}$

Combine like terms and solve for $n$.

Q16. Solve for $r$: $10^{3 - 2r} \cdot 18^{1 - 7r} = 73$

Background

Topic: Exponential Equations (Logarithms)

This question tests your ability to use logarithms to solve for a variable in an equation with multiple exponential terms.

Key Terms and Formulas:

  • Logarithm: $\log(a \cdot b) = \log(a) + \log(b)$

Step-by-Step Guidance

  1. Take the natural logarithm (or log base 10) of both sides: $\ln(10^{3-2r} \cdot 18^{1-7r}) = \ln(73)$.

  2. Use properties of logarithms: $\ln(10^{3-2r}) + \ln(18^{1-7r}) = \ln(73)$.

  3. Bring exponents down: $(3-2r)\ln(10) + (1-7r)\ln(18) = \ln(73)$.

  4. Expand and collect like terms in $r$.

Try solving on your own before revealing the answer!

Final Answer: $r = \frac{3\ln(10) + 1\ln(18) - \ln(73)}{2\ln(10) + 7\ln(18)}$

Combine like terms and solve for $r$.

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