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vfiekz [6]3 years ago

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Step-by-step explanation:

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Answer:

$log_3_2(5)=\frac{1}{5} k$

Step-by-step explanation:

Let's start by using change of base property:

$log_b(x)=\frac{log_a(x)}{log_a(b)}$

So, for $log_2(5)$

$log_2(5)=k=\frac{log(5)}{log(2)}\hspace{10}(1)$

Now, using change of base for $log_3_2(5)$

$log_3_2(5)=\frac{log(5)}{log(32)}$

You can express $32$ as:

$2^5$

Using reduction of power property:

$log_z(x^y)=ylog_z(x)$

$log(32)=log(2^5)=5log(2)$

Therefore:

$log_3_2(5)=\frac{log(5)}{5*log(2)}=\frac{1}{5} \frac{log(5)}{log(2)}\hspace{10}(2)$

As you can see the only difference between (1) and (2) is the coefficient $\frac{1}{5}$ :

So:

$\frac{log(5)}{log(2)} =k\\$

$log_3_2(5)=\frac{1}{5} \frac{log(5)}{log(2)} =\frac{1}{5} k$

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Step-by-step explanation:

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