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 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.`
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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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