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weqwewe [10]
3 years ago
15

If log2 5 = k, determine an expression for log32 5 in terms of k.

Mathematics
1 answer:
lukranit [14]3 years ago
6 0

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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