Logarithmic and Exponential Equations – Questions with Solutions

Solution A: A fractional exponent of \( 0.5 \) is a square root.

\[ 10^{4^{0.5}} = 10^{\sqrt{4}} = 10^2 = 100 \]

Answer: \( \boxed{100} \)

Solution B: Apply the power rule of logarithms to move the multiplier \( \frac{1}{2} \) into the exponent.

\[ 10^{\frac{\log(9)}{2}} = 10^{\frac{1}{2}\log(9)} = 10^{\log(9^{1/2})} = 10^{\log(3)} = 3 \]

Answer: \( \boxed{3} \)

Solution C: Move the coefficient \( 3 \) into the exponent before cancelling \( e \) and \( \ln \).

\[ e^{3\ln(4)} = e^{\ln(4^3)} = 4^3 = 64 \]

Answer: \( \boxed{64} \)

Solution D: Apply the power rule for logarithms.

\[ 10^{-2\log(4)} = 10^{\log(4^{-2})} = 4^{-2} = \frac{1}{16} \]

Answer: \( \boxed{\frac{1}{16}} \)

Solution E: Simplify the exponent using logarithm properties.

\[ e^{2\ln(4)-\ln(16)} = e^{\ln(4^2)-\ln(16)} = e^{\ln(16)-\ln(16)} = e^0 = 1 \]

Answer: \( \boxed{1} \)

Solution A: Recognize that 3 is the square root of 9.

\[ \log_{9}(3) = \log_{9}(9^{1/2}) = \frac{1}{2} \]

Answer: \( \boxed{\dfrac{1}{2}} \)

Solution B: Express the fraction as a negative power of the base.

\[ \log_{2}\left(\frac{1}{8}\right) = \log_{2}(2^{-3}) = -3 \]

Answer: \( \boxed{-3} \)

Solution C: Use the quotient rule for logarithms: \( \log_b(m) - \log_b(n) = \log_b\!\left(\frac{m}{n}\right) \).

\[ \log_{2}(400) - \log_{2}(100) = \log_{2}\!\left(\frac{400}{100}\right) = \log_{2}(4) = 2 \]

Answer: \( \boxed{2} \)

Solution D: Set the expression equal to \( x \) and solve the resulting exponential equation.

\[ x = \log_{8}(4) \implies 8^x = 4 \] \[ (2^3)^x = 2^2 \implies 2^{3x} = 2^2 \] \[ 3x = 2 \implies x = \frac{2}{3} \]

Answer: \( \boxed{\dfrac{2}{3}} \)

Solution E: Use the quotient rule for logarithms.

\[ \log_{5}(600) - \log_{5}(24) = \log_{5}\!\left(\frac{600}{24}\right) = \log_{5}(25) = 2 \]

Answer: \( \boxed{2} \)

Solution A: Take the natural logarithm \( \ln \) of both sides.

\[ e^x = 4 \implies \ln(e^x) = \ln(4) \implies x = \ln(4) \]

Answer: \( \boxed{x = \ln(4)} \)

Solution B: Take the natural logarithm of both sides.

\[ e^x = \pi^2 \implies \ln(e^x) = \ln(\pi^2) \implies x = 2\ln(\pi) \]

Answer: \( \boxed{x = 2\ln(\pi)} \)

Solution C: A base raised to its own logarithm cancels out.

\[ 4^{\log_{4}(3x-2)} = 10 \implies 3x - 2 = 10 \] \[ 3x = 12 \implies x = 4 \]

Answer: \( \boxed{x = 4} \)

Check: \( 3(4) - 2 = 10 \gt 0 \), so the argument of the logarithm is valid. ✓

Solution D: The base 10 and \( \log_{10} \) cancel out.

\[ 10^{\log_{10}(3x)} = 15 \implies 3x = 15 \implies x = 5 \]

Answer: \( \boxed{x = 5} \)

Solution E: Factor out the term with the smallest exponent, \( 3^{x-1} \).

\[ 3^{x-1}(3^2 + 3^1 + 1) = 39 \] \[ 3^{x-1}(9 + 3 + 1) = 39 \] \[ 3^{x-1}(13) = 39 \] \[ 3^{x-1} = 3^1 \implies x - 1 = 1 \implies x = 2 \]

Answer: \( \boxed{x = 2} \)

Solution A: Convert from logarithmic to exponential form twice.

\[ \ln(\ln x) = 4 \implies \ln x = e^4 \implies x = e^{e^4} \]

Answer: \( \boxed{x = e^{e^4}} \)

Solution B: Use logarithmic rules to condense both sides.

\[ \ln\!\left(\frac{x}{4}\right) = \ln\!\left(\frac{x^2}{16}\right) \]

Drop the natural logarithms and solve the resulting equation:

\[ \frac{x}{4} = \frac{x^2}{16} \implies 16x = 4x^2 \implies 4x^2 - 16x = 0 \] \[ 4x(x - 4) = 0 \]

Since the domain of a logarithm requires \( x \gt 0 \), \( x = 0 \) is invalid. Thus \( x = 4 \).

Answer: \( \boxed{x = 4} \)

Solution C: Simplify the right side before taking the fifth root.

\[ (\ln x)^5 = (25^{1/2})^5 = 5^5 \] \[ \ln x = 5 \implies x = e^5 \]

Answer: \( \boxed{x = e^5} \)

Solution D: Condense the right side and drop the logarithms.

\[ \log(x-1) = \log\!\left(\frac{6}{x}\right) \implies x - 1 = \frac{6}{x} \] \[ x^2 - x - 6 = 0 \implies (x - 3)(x + 2) = 0 \]

The domain of \( \log(x-1) \) requires \( x \gt 1 \), making \( x = -2 \) extraneous. Thus \( x = 3 \).

Answer: \( \boxed{x = 3} \)

Solution E: Convert to exponential form.

\[ 3^{x+1} - 18 = 3^2 = 9 \] \[ 3^{x+1} = 27 = 3^3 \implies x + 1 = 3 \implies x = 2 \]

Answer: \( \boxed{x = 2} \)

Solution F: Use the change-of-base identity \( \log_x(7) = \dfrac{1}{\log_7(x)} \). Let \( u = \log_7(x) \).

\[ u + \frac{1}{u} = 2 \implies u^2 - 2u + 1 = 0 \] \[ (u - 1)^2 = 0 \implies u = 1 \]

Substitute back: \( \log_7(x) = 1 \implies x = 7 \).

Answer: \( \boxed{x = 7} \)

Solution G: Convert to exponential form.

\[ x^{3/4} = 27 \]

Raise both sides to the power \( 4/3 \):

\[ x = 27^{4/3} = (\sqrt[3]{27})^4 = 3^4 = 81 \]

Answer: \( \boxed{x = 81} \)