We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

- 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