Question

How do you solve `6^ { x } = 3660`?

Answer 1

`x~~4.579...`

Explanation:

IMPORTANT:
`loga^b=bloga`, as long as base of `log` stays the same

If that 6 on the left side was a 10, we could just take `log_10` of both sides (which is already on standard calculators). However, we can't use a calculator to directly compute `log_6` of anything. So we use what we can and take `log_10` of both sides.

`6^x = 3660`
`log_(10)6^x=log_(10)3660`
`xlog_(10)6=log_(10)3660`
`x=log_(10)3660/log_(10)6`

And because we've isolated `x` on one side, as well as made all `log`s base `10`, you can type in into your calculator:

`x~~4.579...`

Of course you could always put this result back into the original equation, and you should get very close to `3660` (the more decimal places you put back in, the closer you'll get)