How do you solve #(1/5)^x=10#?

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Answer 1 · 2015-09-10T04:15:56

#x=log10/(log1-log5)#

By taking logarithms in both sides we get

#log(1/5)^x=log10=>x(log1-log5)=log10=>x=log10/(log1-log5)#

Answer 2 · 2015-09-10T12:02:44

Use properties of exponents and logs to reformulate and solve, finding:

#x = -1/log(5)#

#(1/5)^x = 5^(-x)#

So #log((1/5)^x) = log(5^(-x)) = -x*log(5)#

#log(10) = 1#

So taking logs of both sides, the original equation becomes:

#-x*log(5) = 1#

Divide both sides by #-log(5)# to get:

#x = -1/log(5)#