# Exponential equation

• Oct 7th 2010, 03:05 PM
matancs92
Exponential equation
Hi guys

I can't solve the following exponential equation:

n * 11^n = 132

Any help will be appreciated!

Matan
• Oct 7th 2010, 05:36 PM
skeeter
Quote:

Originally Posted by matancs92
Hi guys

I can't solve the following exponential equation:

n * 11^n = 132

Any help will be appreciated!

Matan

you'll need to use technology (a calculator or computer) to solve this equation ... $\displaystyle n \approx 1.7928...$
• Oct 7th 2010, 07:00 PM
Educated
$\displaystyle n11^n = 132$

You cannot solve this normally, or get the exact result of it. You can only get an approximation of it (down to however many decimal places you want.)

A method that can be used to solve this is the Newton–Raphson method. $\displaystyle x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

Rearrange the formula to make one side equal zero, $\displaystyle n11^n -132= 0$ and call this f(x).
Select an appropriate number to start off with, call this $\displaystyle x_0$ and apply the newton method.
• Oct 8th 2010, 04:24 AM
HallsofIvy
You can give an "algebraic" solution in terms of the Lambert W function, W(x), which is defined as the inverse function to $\displaystyle f(x)= xe^x$.

Of course, $\displaystyle 11^n= e^{ln(11^n)}= e^{n ln(11)}$. If we let $\displaystyle y= n ln(11)$, then $\displaystyle n= \frac{y}{ln(11)}$ and the equation becomes $\displaystyle \frac{ye^y}{ln(11)}= 132$.

Then $\displaystyle ye^y= 132ln(11)$ so $\displaystyle y= n ln(11)= W(132 ln(11))$ and $\displaystyle n= \frac{W(132 ln(11))}{ln(11)}$.
• Oct 9th 2010, 05:47 AM
matancs92
Thank you very much for helping me guys! :)