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April 28th, 2014, 01:00 PM   #1
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Find the inverse

Find the inverse of the exponential function:

$\displaystyle y={5^{1+x}+1\over 5^x}$

I tried a couple of methods. Here's what I thought to do first. I'll skip a step and switch x and y. Next, I'll multiply the LHS by $\displaystyle 5^y$, giving me:

$\displaystyle 5^y\cdot x=5^{1+y}+1$

I got stuck here. I wasn't sure if I should subtract 1 or separate the $\displaystyle 5^{1+y}$ into $\displaystyle 5\cdot 5^y$. I'm probably just making this problem seem a lot harder than it is. The answer is $\displaystyle y^{-1}=log_{1\over 5}(x-5)$.

Any ideas?
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April 28th, 2014, 01:46 PM   #2
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y = 5 + (1/(5^x))
5^x)= ?
Then, use log.
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April 28th, 2014, 04:15 PM   #3
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Huh? I'm confused.
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April 28th, 2014, 04:50 PM   #4
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$\displaystyle y={5^{1+x}+1\over 5^x}=5+\frac{1}{5^x}$

$\displaystyle y-5=\left(\frac15\right)^x$

$\displaystyle \log_{\frac15}(y-5)=x\Rightarrow y^{-1}=\log_{\frac15}(x-5)$
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April 28th, 2014, 04:59 PM   #5
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Hello, scomora!

Quote:
Find the inverse of the exponential function:

$\displaystyle \qquad y\:=\:\frac{5^{1+x}+1}{5^x}$

Switch variables: $\displaystyle \:x \:=\:\frac{5^{1+y} + 1}{5^y}$

Solve for $\displaystyle y:$

$\displaystyle \qquad\begin{array}{cccc} x\!\cdot\!5^y \:=\:5^{1+y} + 1 \\ \\ x\!\cdot\!5^y \:=\:5\!\cdot\!5^y + 1 \\ \\ x\!\cdot\!5^y - 5\!\cdot\!5^y \:=\:1 \\ \\ (x - 5)5^y \:=\:1 \\ \\ 5^y \:=\:\frac{1}{x-5} \\ \\
5^y \:=\:(x-5)^{\text{-}1}\end{array}$


$\displaystyle \begin{array}{ccc}\text{Take logs: }& \ln(5^y) \:=\:\ln(x-5)^{\text{-}1} \\ \\ & y\!\cdot\!\ln 5 \:=\:-\ln(x-5) \\ \\ & y \:=\:-\frac{\ln(x-5)}{\ln5} \end{array}$

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April 28th, 2014, 10:06 PM   #6
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Quote:
=scomora;192126]Find the inverse of the exponential function:
$\displaystyle y={5^{1+x}+1\over 5^x}$Any ideas?
$\displaystyle y=\frac{5^{1+x}+1}{5^x}=5+5^{-x}$
$\displaystyle y-5=5^{-x}$
$\displaystyle \log_5(y-5)=-x$
$\displaystyle x=-\log_5(y-5)$
$\displaystyle y^{-1}=-\log_5(x-5)$.
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