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Logarithms Practice Test 2

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Question : 8 of 39

Marks: +1, -0

If

\log_2 = 0.3010

, then find the value of

\log_5 512

.

1.876
2.876
3.876
4.876

Solution:

Concept:

Use the change-of-base formula to express

\log_5 512

in terms of

\log_{10} 2

.

Explanation:

Given

\log_{10} 2 = 0.3010

.

By change-of-base,

\log_5 512 = \dfrac{\log_{10} 512}{\log_{10} 5}

.

Write

512 = 2^9

, so

\log_{10} 512 = 9\log_{10} 2

.

Also,

\log_{10} 5 = \log_{10}(10/2) = \log_{10}10 - \log_{10}2 = 1 - 0.3010

.

Substitute:

\log_5 512 = \dfrac{9 \times 0.3010}{1 - 0.3010}

.

Compute denominator:

1 - 0.3010 = 0.6990

.

Numerator:

9 \times 0.3010 = 2.7090

.

So,

\log_5 512 = \dfrac{2.7090}{0.6990} \approx 3.876

.

Answer:

\log_5 512 \approx 3.876

, so the correct option is C.

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