# Math Help - Help for exponential solving

1. ## Help for exponential solving

Hello! i've already solved a part of it;
So 22-5x=4x
I then got 2-5x=2x
This is supposed to give 2/7 as an answer but i always get -5 -_-
Help!!

2. ## Re: Help for exponential solving

Originally Posted by laurettarosa
Hello! i've already solved a part of it;
So 22-5x=4x
I then got 2-5x=2x
This is supposed to give 2/7 as an answer but i always get -5 -_-
Help!!
You started correctly: well done.

$2^{(2 - 5x)} = 4^x \implies 2^{(2 - 5x)} = \left(2^2\right)^x \implies 2^{(2 - 5x)} = 2^{2x} \implies 2 - 5x = 2x.$ That was the hard part. Then

$2 - 5x = 2x \implies 2 - 5x + 5x = 2x + 5x \implies 2 = 2x + 5x \implies 2 = 7x \implies \dfrac{1}{7} * 7x = \dfrac{1}{7} * 2 \implies x = \dfrac{2}{7}.$

Check

$2^{(2 - 5 * \frac{2}{7})} = 2^{\frac{14}{7} - \frac{10}{7}} = 2^{\frac{4}{7}} = \sqrt[7]{2^4} = \sqrt[7]{16} = \sqrt[7]{4^2} = 4^{\frac{2}{7}}.$