# Log/Exponential problem

• Apr 18th 2010, 08:24 PM
desiderius1
Log/Exponential problem
$
\frac{e^x-e^{-x}}{2}=10
$

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

What I'm doing (More than likely wrong)

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

$

$
e^x-e=10(2^x)
$

$
e^x-e=20^x
$
• Apr 18th 2010, 08:33 PM
sa-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 PM
Failure
Quote:

Originally Posted by desiderius1
$
\frac{e^x-{\color{blue}e^{-x}}}{{\color{red}2}}=7
$

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

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)

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

The above equation is very different from the one that you were given.

Quote:

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

$
e^x-e=14^x
$

That last step is not valid. More generally, you have that in general $a\cdot b^n\neq (a\cdot b)^n$.
• Apr 18th 2010, 08:58 PM
desiderius1
Sorry, there are two identical problems, one equating to 10 and the other 7, I misread while typing.

The problem is:
$
\frac{e^x-e^{-x}}{2}=10
$
• Apr 19th 2010, 05:03 AM
Failure
Quote:

Originally Posted by desiderius1
Sorry, there are two identical problems, one equating to 10 and the other 7, I misread while typing.

The problem is:
$
\frac{e^x-e^{-x}}{2}=10
$

The easy way to solve this equation would be to recognize that the left side really is $\sinh(x)$, and that the solution is simply $x=\arsinh(10)$.

Or else you can first multiply the equation by $2e^x$ to get a quadratic equation for $e^x$:

$\frac{e^x-e^{-x}}{2}=10 \qquad | \cdot 2e^x$

$\left(e^x\right)^2-1=20 e^x \qquad | -20e^x$

$\left(e^x\right)^2-20e^x-1=0$
Thus, we get

$\left(e^x\right)_{1,2} = \frac{-(-20)\pm \sqrt{(-20)^2-4\cdot 1\cdot (-1)}}{2\cdot 1}=10\pm \sqrt{101}$
But only the $+$ case is possible, thus we have

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

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