Question

How do you solve `5^(x+2) = 2`?

Answer 1

`x=(log(2/25))/(log5)=-1,57`

Explanation:

From the laws of exponents we may write this as

`5^x*5^2=2`

Now dividing throughout by `5^2` yields

`5^x=2/25`

Now taking the log on both sides and using laws of logs we can write it as

`xlog5=log(2/25)`

Finally dividing throughout by `log5` gives

`x=(log(2/25))/(log5)=-1,57`

Answer 2

To undo the `x` being raised to an exponent, take the log base 5 of both sides. This undoes the 5 on the left, leaving just `x+2`. On the right is `log_"5"2`, which can be put into a calculator using the change of base formula, in which `log_"5"2=(log2)/(log5)`.

Then, 2 is subtracted from both sides, leaving `x` equal to `(log2)/(log5)-2~~-1.57`.