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...

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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.` **