# 25^(x+2)+5=6*5^x+2 How do you solve these equations when you have to multiply the numbers on one side and add them on the other? Please explain!

You need to move all terms to the left side such that:

`25^(x+2)+5=6*5^(x+2)`

`25^(x+2)-6*5^(x+2)+5 = 0`

Since `25 = 5^2` , you may use `5^2`  such that:

`(5^2)^(x+2)-6*5^(x+2)+5 = 0`

Using the exponential identity `(a^x)^y = a^(x*y)`  yields:

`(5^2)^(x+2) = 5^(2(x+2))`

You should come up with the following substitution such...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

You need to move all terms to the left side such that:

`25^(x+2)+5=6*5^(x+2)`

`25^(x+2)-6*5^(x+2)+5 = 0`

Since `25 = 5^2` , you may use `5^2`  such that:

`(5^2)^(x+2)-6*5^(x+2)+5 = 0`

Using the exponential identity `(a^x)^y = a^(x*y)`  yields:

`(5^2)^(x+2) = 5^(2(x+2))`

You should come up with the following substitution such that:

`5^(x+2) = t => 5^(2(x+2)) = t^2`

Changing the variable in equation yields:

`t^2 - 6t + 5 = 0`

`t_(1,2) = (6+-sqrt(36 - 20))/2 => t_(1,2) = (6+-sqrt16)/2`

`t_(1,2) = (6+-4)/2 =>t_1 = 5 , t_2 = 1`

You need to solve for x the following equations such that:

`5^(x+2) = t_1 => 5^(x+2) = 5 => x+2 = 1 => x = -1`

`5^(x+2) = t_2 => 5^(x+2) =1 => 5^(x+2) =5^0 => x+2 = 0 => x = -2`

Hence, evaluating the solutions to the given exponential equation yields `x = -2`  and `x = -1.`

Approved by eNotes Editorial Team