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I have to find $x$ in this equation. I have already tried cross multiplying but I am not sure what to do after. $$ \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{3}{4} $$

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    Do you know about the hyperbolic functions? – 2017-01-23
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    @pie314271 not yet. we are still in logarithmic functions in class. – 2017-01-23
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    Format your questions properly! – 2017-01-23
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    @tired sorry i will try fixing it. the formatting wont show up earlier :( – 2017-01-23
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    Hint set $z=e^x$ and solve the resulting quadratic equation – 2017-01-23
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    Yes, in that case you would just use tired's method. – 2017-01-23

3 Answers 3

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Let $z=e^x$. Then

$$\frac{z-z^{-1}}{z+z^{-1}}=\frac34,$$

which leads to the quadratic equation

$$4(z^2-1)=3(z^2+1),$$

or

$$z^2=7.$$

Hence $x=\dfrac{\log7}2$, which is the only solution.

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    (Independently of @tired). – 2017-01-23
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    https://www.quora.com/What-is-the-componendo-and-dividendo-rule – 2017-01-23
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We know that $$\sinh x =\frac {e^x -e^{-x}}{2} $$ and that $$\cosh x =\frac {e^x + e^{-x}}{2} $$ Can you now proceed where $\sinh x $ and $\cosh x $ are hyperbolic trigonometric functions?

We can otherwise put $k =e^x $ and then get our equation as $$\frac {k-\frac {1}{k}}{k+\frac {1}{k}} =\frac {3}{4} $$ $$4 (k^2-1) =3 (k^2+1) $$ $$\Rightarrow k = \sqrt {7} \text { as an acceptable solution} $$ Hope it helps.

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    Isn't it simpler to just state that the equation is $\tanh(x)=3/4$ ? – 2017-01-23
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Here is a solution without the use of hyperbolic trigonometric functions.

We have:

$$\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{3}{4}$$

Let's multiply both the numerator and denominator of the left hand side by $e^x$: $$\frac{(e^x-e^{-x})\cdot e^x}{(e^x+e^{-x})\cdot e^x}=\frac{3}{4}$$ $$\frac{e^{2x}-1}{e^{2x}+1}=\frac{3}{4}$$

Now, you can substitute $u=e^{2x}$, and obtain a linear equation after doing cross-multiplication: $$\frac{u-1}{u+1}=\frac{3}{4}$$ You can now easily solve for $u$, and substitute back $u=e^{2x}$. Then, solve for $x$.

If you have any questions, please do not hesitate to ask.