Question

1) Log(4)64=x 2) Log(3)243=2x+1 The numbers in parenthesis is the base. I know the first one I can just throw in the calculator, but that's now how we're supposed to solve these ones. What's the value of x?

Answer 1

Have a look:

Explanation:

In the first you do not need the calculator.
Consider:
`log_4(64)=x`
use the definition of log...what is the number `x` you need to raise `x` to get `64`?
or: `4^x=64`
you can see that is `3` in fact `4^3=64`
so `x=3`
Also the second do not require the calculator in fact you know that `3^5=243` so you can write:
`log_3(3^5)=2x+1` use again the definition of log:
`3^5=3^(2x+1)` so that:
`5=2x+1`
`x=4/2=2`