How do you solve #5^-x = 250#?

1 Answer
KillerBunny
Oct 24, 2015

#x=-log_5(250)#

Explanation:

Since the logarithm is the inverse function of the exponential (i.e., #log_a(a^x)=x#, you can use #log_5# to isolate the #x#:

#5^{-x}=250 \implies log_5(5^{-x})=log_5(250)#,

but #log_5(5^{-x})=-x#.

So, the equation becomes #-x=log_5(250)#, which we easily solve for #x# changing the sign.

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