Question

How do you solve for x in `log_x(32)=5`?

Answer 1

If `log_x(32)=5` then `x^5 = x^(log_x(32)) = 32`.

So `x = root(5)(32) = root(5)(2^5) = 2`.

Explanation:

Alternatively, use the change of base formula:

`5 = log_x(32) = ln(32)/ln(x)`

So `ln(x) = ln(32)/5 = ln(root(5)(32)) = ln(root(5)(2^5)) = ln(2)`

Hence `x = 2`