Question

How do you solve `3.14159^x=4`?

Answer 1

`x=ln(4)/ln(3.14159)=log_3.14159(4)`

Explanation:

We will use the property of logarithms that `ln(a^x) = xln(a)`

With that:

`3.14159^x = 4`

`=> ln(3.14159^x)=ln(4)`

`=> xln(3.14159)=ln(4)`

`:. x=ln(4)/ln(3.14159)`

Note that this is eqivalent to the base `3.14159` log of `4`, a result we could have also found by taking the base `3.14159` log of both sides and applying `log_a(a^x)=x`