$\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

$

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- Apr 18th 2010, 08:24 PMdesiderius1Log/Exponential problem
$\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

$ - Apr 18th 2010, 08:33 PMsa-ri-ga-ma
According to the answer, the given problem is wrong.

The equation should be equated to 10 rather than 7. - Apr 18th 2010, 08:37 PMFailure
This answer is wrong, assuming that you have managed to write down the correct equation that you were asked to solve.

Quote:

What I'm doing (More than likely wrong)

$\displaystyle

\frac{e^x - {\color{blue}e}}{{\color{red}2^x}} = 7

$

Quote:

$\displaystyle

e^x-e=7(2^x)

$

$\displaystyle

e^x-e=14^x

$

- Apr 18th 2010, 08:58 PMdesiderius1
Sorry, there are two identical problems, one equating to 10 and the other 7, I misread while typing.

The problem is:

$\displaystyle

\frac{e^x-e^{-x}}{2}=10

$ - Apr 19th 2010, 05:03 AMFailure
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

$\displaystyle e^x=10+\sqrt{101}\qquad | \ln$

$\displaystyle x=\ln(10+\sqrt{101})$