# 25^(x+2)+5=6*5^x+2How 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 that:

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

Changing the variable in equation yields:

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

Using quadratic formula yields:

`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.` **

I assume the equation is: 25^(x+2)+5=6*5^(x+2).

What we need to do is notice that we can write first write the equation in terms of y = 5^(x+2).

25^(x+2) = (5^2)^(x+2) = 5^(2*(x+2)) = (5^(x+2))^2 = y^2

and

6*5^(x + 2) = 6*y

so the equation can be written as

y^2 + 5 = 6*y

This is just a quadratic equation

y^2 - 6*y + 5 = (y-5)(y-1)

So y=5 or y=1

which means 5^(x+2) = 5 or 5^(x+2) = 1

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

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

Hence the answer is x=-1 or x=-2.