How do you solve #25^x=1/125# ?

2 Answers

#x= -3/2#

#x=-1.5#

Explanation:

#1/125 = 5^-3#

so #25^x = 5^-3#

and #(5^2)^x = 5^-3#

so #5^(2x) = 5^-3#

and now that the bases are the same you can remove them:

#2x = -3#

#x= -3/2#

#x=-1.5#

Mar 7, 2018

Here we might need a common base, or you can use log and a calculator
#x= -3/2 or -1.5#

Explanation:

Method 1- Common base
I noticed that both 25 and 125 are 5 to some power.
So therefore I can safely state that:
#5^(2x)= 1/5^3#
#5^(2x)= 5^(-3)#
Now that we have a common base of 5, I can just set the exponents equal to one another.
#2x=-3#
#x=-3/2#

Method 2- Use log
#25^x=1/125# is the same thing as #log_25(1/125)=x#
#"plug" log_25(1/125) "into your calculator and you should still get" -1.5#