Test Index

Logarithms

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Question : 49 of 85

Marks: +1, -0

If

\log_{e} 2 \log_{b} 625 = \log_{10} 16 \cdot \log_{e} 10,

b=

4
5
1
e.

Solution:

Concept:

Convert all logarithms to natural logs and simplify.

Explanation:

Given:

\ln 2 \cdot \log_b 625 = \log_{10} 16 \cdot \ln 10

.

Rewrite

\log_b 625 = \frac{\ln 625}{\ln b}

and

\log_{10} 16 = \frac{\ln 16}{\ln 10}

.

Right side becomes

\frac{\ln 16}{\ln 10} \cdot \ln 10 = \ln 16

.

So

\ln 2 \cdot \frac{\ln 625}{\ln b} = \ln 16

.

Now

\ln 625 = \ln(5^4) = 4\ln 5

, and

\ln 16 = \ln(2^4) = 4\ln 2

.

Thus

\ln 2 \cdot \frac{4\ln 5}{\ln b} = 4\ln 2

.

Cancel

\ln 2

(non-zero):

\frac{4\ln 5}{\ln b} = 4

.

Multiply by

\ln b

:

4\ln 5 = 4\ln b

, so

\ln 5 = \ln b

.

Hence

b = 5

.

Answer:

5

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