How do you solve #5^x = 71#?

1 Answer
sente
Feb 26, 2016

#x=log_5(71)#

Explanation:

We will be using the following properties of logarithms:

  • #log(a^x) = xlog(a)#

  • #log_a(a) = 1#

Then, to solve, we simply take the base-5 logarithm of each side of the equation.

#5^x = 71#

#=>log_5(5^x) = log_5(71)#

#=>xlog_5(5)=log_5(71)#

#=>x*1=log_5(71)#

#=>x=log_5(71)#

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