Question

question 5
evaluate the following logarithms without a calculator (no decimal answers):
\\(\log_{125}(25) = \\)
\\(\log_{16}(64) = \\)
\\(\log_{27}(81) = \\)
question help: video

Explanation:

Step1: Recall logarithm definition

Let \( \log_{a}(b) = x \), then \( a^x = b \). For \( \log_{125}(25) \), let \( x = \log_{125}(25) \), so \( 125^x = 25 \). Since \( 125 = 5^3 \) and \( 25 = 5^2 \), we have \( (5^3)^x = 5^2 \), which simplifies to \( 5^{3x} = 5^2 \). Thus, \( 3x = 2 \), so \( x = \frac{2}{3} \).

Step2: Evaluate \( \log_{16}(64) \)

Let \( y = \log_{16}(64) \), so \( 16^y = 64 \). \( 16 = 2^4 \) and \( 64 = 2^6 \), so \( (2^4)^y = 2^6 \), or \( 2^{4y} = 2^6 \). Then \( 4y = 6 \), so \( y = \frac{6}{4} = \frac{3}{2} \).

Step3: Evaluate \( \log_{27}(81) \)

Let \( z = \log_{27}(81) \), so \( 27^z = 81 \). \( 27 = 3^3 \) and \( 81 = 3^4 \), so \( (3^3)^z = 3^4 \), or \( 3^{3z} = 3^4 \). Then \( 3z = 4 \), so \( z = \frac{4}{3} \).

Answer:

\( \log_{125}(25) = \frac{2}{3} \), \( \log_{16}(64) = \frac{3}{2} \), \( \log_{27}(81) = \frac{4}{3} \)