How do you solve $5^{x} = 71$ $5^x = 71$ ?

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Answer 1 · 2016-02-26T02:46:14

$x = {log}_{5} (71)$ $x=log_5(71)$

We will be using the following properties of logarithms:

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

$5^{x} = 71$ $5^x = 71$

$\Rightarrow {log}_{5} (5^{x}) = {log}_{5} (71)$ $=>log_5(5^x) = log_5(71)$

$\Rightarrow x {log}_{5} (5) = {log}_{5} (71)$ $=>xlog_5(5)=log_5(71)$

$\Rightarrow x \cdot 1 = {log}_{5} (71)$ $=>x*1=log_5(71)$

$\Rightarrow x = {log}_{5} (71)$ $=>x=log_5(71)$

How do you solve 5x=715^x = 71?