How do you solve #3^ { x } = \sqrt { 5^ { x - 2} }#?

1 Answer
May 25, 2017

#x=log_(5/9)(25)#

Explanation:

Let's rewrite the equation:
#3^x=(5^(x-2))^(1/2)#

#(3^x)^2=(5^(x-2))^(1/2*2)#
#3^(2x)=(5^x)/5^2#
#(5^x)/3^(2x)=25#
#(5/9)^x=25#

#x=log_(5/9)(25)~~-5.48#

If you are unfamiliar with the concept of logarithm, here's a quick overview:

If we have #b^x=y#
Then, #x=log_(b)(y)#, pronounced "log base b of y."