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!

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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...

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widowspeak | Student

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.

 

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