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Logarithms Practice Test 2

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Question : 13 of 39

Marks: +1, -0

\frac{\log_8 17}{\log_9 23} - \frac{\log_{2\sqrt{2}} 17}{\log_3 23} = ?

0
1
17/8
23/17

Solution:

Concept:

Convert the logarithms to a common base and simplify.

Explanation:

Write

\log_8 17

in base

2

.

\log_8 17=\frac{\log_2 17}{\log_2 8}=\frac{\log_2 17}{3}

.

Since

2\sqrt{2}=2^{3/2}

,

\log_{2\sqrt{2}}17=\frac{\log_2 17}{\log_2 2^{3/2}}=\frac{\log_2 17}{3/2}=\frac{2}{3}\log_2 17

.

Write

\log_9 23

in base

3

.

\log_9 23=\frac{\log_3 23}{\log_3 9}=\frac{\log_3 23}{2}

.

First term:

\frac{\log_8 17}{\log_9 23}=\frac{\frac{1}{3}\log_2 17}{\frac{1}{2}\log_3 23}=\frac{2}{3}\frac{\log_2 17}{\log_3 23}

.

Second term:

\frac{\log_{2\sqrt{2}}17}{\log_3 23}=\frac{\frac{2}{3}\log_2 17}{\log_3 23}=\frac{2}{3}\frac{\log_2 17}{\log_3 23}

.

Both terms are equal, so their difference is

0

.

Answer:

0

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