# Extraneous solutions, absolute value

• Jun 22nd 2011, 06:26 AM
Zellator
Extraneous solutions, absolute value
Hi Forum!
I've just found something strange

The closest integral of sum of the solutions to $\displaystyle 27^{5x+3}=81^{|x+2|-7}$ is

OK, calculating, we get two solutions

$\displaystyle \frac{-29}{11}$ and $\displaystyle \frac{-45}{19}$

But the solution says that $\displaystyle \frac{-29}{11}$ should be disconsidered.
Why?
The sum of them would give something about $\displaystyle -5$.
Can someone explain?
Does $\displaystyle x=\frac{-29}{11}$ makes the function diverge?

Thanks
• Jun 22nd 2011, 06:46 AM
Prove It
Re: Extraneous solutions, absolute value
Does $\displaystyle \displaystyle 27^{5\left(-\frac{29}{11}\right)+3}$ equal $\displaystyle \displaystyle 81^{\left|-\frac{29}{11} + 2\right| - 7}$?
• Jun 22nd 2011, 07:59 AM
Zellator
Re: Extraneous solutions, absolute value
Quote:

Originally Posted by Prove It
Does $\displaystyle \displaystyle 27^{5\left(-\frac{29}{11}\right)+3}$ equal $\displaystyle \displaystyle 81^{\left|-\frac{29}{11} + 2\right| - 7}$?

Hi Prove It
I get your point, then it is necessary to test the values.
I thought there was something else to it; this is really simple. :)