How do you solve for x in ${log}_{x} 32 = 5$ $log_x32=5$ ?

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Answer 1 · 2018-04-26T12:37:13

I tried this:

I would try to use the definition of log to write it as:

$x^{5} = 32$ $x^5=32$

now let us observe $32$ $32$ ; it can be written as: $2^{5}$ $2^5$

so we can write:

$x^{5} = 2^{5}$ $x^5=2^5$

and so $x = 2$ $x=2$

How do you solve for x in logx32=5log_x32=5?