How do I solve ${ln (\frac{1}{2})}^{m} = ln 2$ $ln(1/2)^m = ln2$ ?

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How do I solve ${ln (\frac{1}{2})}^{m} = ln 2$ $ln(1/2)^m = ln2$ ?

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Answer 1 · 2017-04-01T07:26:25

Two ways...

One is simply recognizing that by algebra, $2^{- 1} = \frac{1}{2}$ $2^(-1) = 1/2$ . Hence, $m = - 1$ $m = -1$ by inspection, i.e. ${(2^{- 1})}^{- 1} = 2 = {(\frac{1}{2})}^{- 1}$ $(2^(-1))^(-1) = 2 = (1/2)^(-1)$ .

Another way is that anytime you see an exponent, try taking the $ln$ $ln$ of both sides.

${ln (\frac{1}{2})}^{m} = ln 2$ $ln(1/2)^(m) = ln2$

Use the property that

${ln a}^{b} = b ln a$ $lna^b = blna$

Thus, we get:

$m ln (\frac{1}{2}) = ln 2$ $mln(1/2) = ln2$

$m {ln (2)}^{- 1} = ln 2$ $mln(2)^(-1) = ln2$

$- m ln 2 = ln 2$ $-mln2 = ln2$

$\Rightarrow m = - 1$ $=> m = -1$

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How do I solve ${ln (\frac{1}{2})}^{m} = ln 2$ $ln(1/2)^m = ln2$ ?

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