# 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!

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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.