- 5
^{16-(x2-x)}= 25^{- (x2-}_{8)} The "2" numbers are exponents (b^{n}) - 25
^{4x }= 1/5^{x}^{2}

^{Please, I really need help.}

chaconsara32
Oct 18, 2017

#1**+2 **

Hard to read the frst one but I assume this is :

1. 5^{16 -(x^2 - x)} = 25^{-(x^2 - 8)} simplify as

5 ^{-x^2 + x + 16} = (5^{2}) ^{8 - x^2}

5 ^{-x^2 + x + 16} = 5^{16 -x^2}

Since the bases are the same.....solve for the exponents

-x^2 + x + 16 = 16 - x^2 subtract x^2 from both sides

x + 16 = 16 subtract 16 from both sides

x = 0

CPhill
Oct 18, 2017

#2**+1 **

Yes, it is like that I still need to get the hang of writing these on a computer! Thanks!

chaconsara32
Oct 18, 2017

#3**+2 **

Here's the second one

25^{4x} = (1/5)^{x^2}

Note that 1/5 = 5^-1 and 25 = 5^2 ....so we have....

(5^2)^{4x} = (5^-1)^{x^2}

5^{8x} = 5^{-x^2} the bases are the same....solve for the exponents

8x = -x^2 add x^2 to both sides

x^2 + 8x = 0 factor

x ( x + 8) = 0 set both factors to 0 and solve for x and we get that

x = 0 or x = -8

CPhill
Oct 18, 2017