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

Re: Extraneous solutions, absolute value

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

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. :)

Thanks for your reply.

All the best.