Question

How do you solve `125= 10^{x}`?

Answer 1

#x=3log5=2.097

Explanation:

Using logs
`10^x=125=5^3`
`log₁₀10^x=log₁₀5^3`
`xlog₁₀10=3log₁₀5`
`x=(3log₁₀5)/(log₁₀10)=3log₁₀5=2.097`