How do you find the inverse of $y = {log}_{\frac{1}{2}} x$ $y=log_(1/2) x$ ?

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Answer 1 · 2016-01-09T19:13:26

It's ${(\frac{1}{2})}^{x}$ $(1/2)^x$

${log}_{\frac{1}{2}} (x) = \frac{ln (x)}{ln (\frac{1}{2})}$ $log_(1/2)(x) = ln(x)/ln(1/2)$

by logic you put ${(\frac{1}{2})}^{x}$ $(1/2)^x$

$\frac{ln ({(\frac{1}{2})}^{x})}{ln (\frac{1}{2})}$ $ln((1/2)^x)/ln(1/2)$ with the identity $a ln (b) = ln (b^{a})$ $aln(b) = ln(b^a)$ you transform

$x \frac{ln (\frac{1}{2})}{ln (\frac{1}{2})} = x$ $xln(1/2)/ln(1/2) = x$

You find the identity application so $y = {(\frac{1}{2})}^{x}$ $y = (1/2)^x$ is the inverse function because

$f \circ f^{- 1} = x$ $f@f^(-1) = x$

Answer 2 · 2016-01-09T20:55:27

$y = {(\frac{1}{2})}^{x}$ $y=(1/2)^x$

Given $color(white)(....)y=log_(1/2)(x)$

By the nature of what a log is, another way of writing the given relationship is:

$(1/2)^y=x$

As it turns out this is exactly the format needed for the inverse function. All you have to now do is swap the variable letters round to give:

So if $f(x)=log_(1/2)(x)$

Then $f^(-)(x) =(1/2)^x$

Or, if you prefer $-> y=(1/2)^x$

How do you find the inverse of y=log_(1/2) xy=log12xy=log_(1/2) x?