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

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Answer 1 · 2015-08-21T21:04:54

If ${log}_{x} (32) = 5$ $log_x(32)=5$ then $x^{5} = x^{{log}_{x} (32)} = 32$ $x^5 = x^(log_x(32)) = 32$ .

So $x = \sqrt[5]{32} = \sqrt[5]{2^{5}} = 2$ $x = root(5)(32) = root(5)(2^5) = 2$ .

Alternatively, use the change of base formula:

$5 = {log}_{x} (32) = \frac{ln (32)}{ln (x)}$ $5 = log_x(32) = ln(32)/ln(x)$

So $ln (x) = \frac{ln (32)}{5} = ln (\sqrt[5]{32}) = ln (\sqrt[5]{2^{5}}) = ln (2)$ $ln(x) = ln(32)/5 = ln(root(5)(32)) = ln(root(5)(2^5)) = ln(2)$

Hence $x = 2$ $x = 2$

How do you solve for x in logx(32)=5log_x(32)=5?