Question

How do you solve `25^x=1/125` ?

Answer 1

`x= -3/2`

`x=-1.5`

Explanation:

`1/125 = 5^-3`

so `25^x = 5^-3`

and `(5^2)^x = 5^-3`

so `5^(2x) = 5^-3`

and now that the bases are the same you can remove them:

`2x = -3`

`x= -3/2`

`x=-1.5`

Answer 2

Here we might need a common base, or you can use log and a calculator
`x= -3/2 or -1.5`

Explanation:

Method 1- Common base
I noticed that both 25 and 125 are 5 to some power.
So therefore I can safely state that:
`5^(2x)= 1/5^3`
`5^(2x)= 5^(-3)`
Now that we have a common base of 5, I can just set the exponents equal to one another.
`2x=-3`
`x=-3/2`

Method 2- Use log
`25^x=1/125` is the same thing as `log_25(1/125)=x`
`"plug" log_25(1/125) "into your calculator and you should still get" -1.5`