$\displaystyle
\frac{e^x-e^{-x}}{2}=10
$
Answer:
$\displaystyle
x = ln (10 + \sqrt{101} )
$
What I'm doing (More than likely wrong)
$\displaystyle
\frac{e^x - e}{2^x} = 10
$
$\displaystyle
e^x-e=10(2^x)
$
$\displaystyle
e^x-e=20^x
$
$\displaystyle
\frac{e^x-e^{-x}}{2}=10
$
Answer:
$\displaystyle
x = ln (10 + \sqrt{101} )
$
What I'm doing (More than likely wrong)
$\displaystyle
\frac{e^x - e}{2^x} = 10
$
$\displaystyle
e^x-e=10(2^x)
$
$\displaystyle
e^x-e=20^x
$
This answer is wrong, assuming that you have managed to write down the correct equation that you were asked to solve.
The above equation is very different from the one that you were given.What I'm doing (More than likely wrong)
$\displaystyle
\frac{e^x - {\color{blue}e}}{{\color{red}2^x}} = 7
$
That last step is not valid. More generally, you have that in general $\displaystyle a\cdot b^n\neq (a\cdot b)^n$.$\displaystyle
e^x-e=7(2^x)
$
$\displaystyle
e^x-e=14^x
$
The easy way to solve this equation would be to recognize that the left side really is $\displaystyle \sinh(x)$, and that the solution is simply $\displaystyle x=\arsinh(10)$.
Or else you can first multiply the equation by $\displaystyle 2e^x$ to get a quadratic equation for $\displaystyle e^x$:
$\displaystyle \frac{e^x-e^{-x}}{2}=10 \qquad | \cdot 2e^x$
$\displaystyle \left(e^x\right)^2-1=20 e^x \qquad | -20e^x$
$\displaystyle \left(e^x\right)^2-20e^x-1=0$
Thus, we get
$\displaystyle \left(e^x\right)_{1,2} = \frac{-(-20)\pm \sqrt{(-20)^2-4\cdot 1\cdot (-1)}}{2\cdot 1}=10\pm \sqrt{101}$
But only the $\displaystyle +$ case is possible, thus we have
$\displaystyle e^x=10+\sqrt{101}\qquad | \ln$
$\displaystyle x=\ln(10+\sqrt{101})$