Question

y=(10^x-10^-x)/(10^x+10^-x)

find x in terms of y

Answer 1

Recall the definitions of the hyperbolic trig functions:
\[\sinh x=\frac{e^x-e^{-x}}{2}~~\text{and}~~\cosh x=\frac{e^x+e^{-x}}{2}\]
Notice that youhave
\[\tanh x=\frac{\sinh x}{\cosh x}=\frac{e^x-e^{-x}}{e^x+e^{-x}}\]
What we want to do is write \(10^x\) and \(10^{-x}\) in terms of the base \(e\), which we can do using some properties of logarithms and exponents:
\[b^{f(x)}=\exp\left(\ln b^{f(x)}\right)=\exp\left(f(x)\ln b\right)=e^{\ln b\,f(x)}\]
In this case, you have \(f(x)=x\) and \(f(x)=-x\) and \(b=10\).

So,
\[\begin{align*}
y&=\frac{10^x-10^{-x}}{10^x+10^{-x}}\\\\
&=\frac{e^{x\ln 10}-e^{-x\ln 10}}{e^{x\ln 10}+e^{-x\ln 10}}\\\\
&=\tanh(x\ln 10)\\\\
&=\tanh\left(\ln10^x\right)\\\\
\tanh^{-1}y&=\ln 10^x\\\\
\exp\left(\tanh^{-1}y\right)&=10^x\\\\
\log_{10}\left[\exp\left(\tanh^{-1}y\right)\right]&=x
\end{align*}\]

Answer 2

Alternate method.





\(\large \begin{align} \color{black}{y=\dfrac{(10^x-10^{-x})}{(10^x+10^{-x})}\hspace{.33em}\\~\\
y+1=\dfrac{(10^x-10^{-x})}{(10^x+10^{-x})}+1\hspace{.33em}\\~\\
y+1=\dfrac{(10^x-10^{-x})+(10^x+10^{-x})}{(10^x+10^{-x})}\hspace{.33em}\\~\\
y+1=\dfrac{(2\times 10^x)}{(10^x+10^{-x})}\hspace{.33em}\\~\\
\dfrac{1}{y+1}=\dfrac{(10^x+10^{-x})}{(2\times 10^x)}\hspace{.33em}\\~\\
\dfrac{2}{y+1}=\dfrac{2(10^x+10^{-x})}{(2\times 10^x)}\hspace{.33em}\\~\\
\dfrac{2}{y+1}-1=\dfrac{2(10^x+10^{-x})}{(2\times 10^x)}-1\hspace{.33em}\\~\\
\dfrac{2-(y+1)}{y+1}=\dfrac{2(10^x+10^{-x})-(2\times 10^x)}{(2\times 10^x)}\hspace{.33em}\\~\\
\dfrac{(1-y)}{y+1}=\dfrac{2\times 10^{-x})}{(2\times 10^x)}\hspace{.33em}\\~\\
\dfrac{(1-y)}{y+1}=\dfrac{( 10^{-x})}{( 10^x)}\hspace{.33em}\\~\\
\dfrac{(1-y)}{y+1}=\dfrac{1}{( 10^{2x})}\hspace{.33em}\\~\\
10^{2x}=\dfrac{y+1}{y-1} \hspace{.33em}\\~\\
2x\log_{10} 10=\log_{10}\left(\dfrac{y+1}{y-1}\right) \hspace{.33em}\\~\\
2x=\log_{10}\left(\dfrac{y+1}{y-1}\right) \hspace{.33em}\\~\\
x=\dfrac{\log_{10}\left(\dfrac{y+1}{y-1}\right)}{2} \hspace{.33em}\\~\\
}\end{align}\)

Answer 3

@mathmath333 in the 12th (I think) line, you made a slight error. The denominator in the logarithm should be \(1-y\).

Answer 4

yes thnkx for pointing ,let me correct it

Answer 5

\(\large \begin{align} \color{black}{y=\dfrac{(10^x-10^{-x})}{(10^x+10^{-x})}\hspace{.33em}\\~\\
y+1=\dfrac{(10^x-10^{-x})}{(10^x+10^{-x})}+1\hspace{.33em}\\~\\
y+1=\dfrac{(10^x-10^{-x})+(10^x+10^{-x})}{(10^x+10^{-x})}\hspace{.33em}\\~\\
y+1=\dfrac{(2\times 10^x)}{(10^x+10^{-x})}\hspace{.33em}\\~\\
\dfrac{1}{y+1}=\dfrac{(10^x+10^{-x})}{(2\times 10^x)}\hspace{.33em}\\~\\
\dfrac{2}{y+1}=\dfrac{2(10^x+10^{-x})}{(2\times 10^x)}\hspace{.33em}\\~\\
\dfrac{2}{y+1}-1=\dfrac{2(10^x+10^{-x})}{(2\times 10^x)}-1\hspace{.33em}\\~\\
\dfrac{2-(y+1)}{y+1}=\dfrac{2(10^x+10^{-x})-(2\times 10^x)}{(2\times 10^x)}\hspace{.33em}\\~\\
\dfrac{(1-y)}{y+1}=\dfrac{2\times 10^{-x})}{(2\times 10^x)}\hspace{.33em}\\~\\
\dfrac{(1-y)}{y+1}=\dfrac{( 10^{-x})}{( 10^x)}\hspace{.33em}\\~\\
\dfrac{(1-y)}{y+1}=\dfrac{1}{( 10^{2x})}\hspace{.33em}\\~\\
10^{2x}=\dfrac{y+1}{1-y} \hspace{.33em}\\~\\
2x\log_{10} 10=\log_{10}\left(\dfrac{y+1}{1-y}\right) \hspace{.33em}\\~\\
2x=\log_{10}\left(\dfrac{y+1}{1-y}\right) \hspace{.33em}\\~\\
x=\dfrac{\log_{10}\left(\dfrac{y+1}{1-y}\right)}{2} \hspace{.33em}\\~\\
}\end{align}\)

Answer 6

Thanks for the replies!

Answer 7

how to know when to add 1 or mulitply 2?