How do you solve this??

Question

(x^2-9x+20)
(x^2-5x+5 ) =1

By the way, (x^2-9x+20) is an exponent of the bottom number.
Can anybody help me solve it...thanks

(x^2-9x+20)
(x^2-5x+5) =1

------------ (x^2-9x+20)
(x^2-5x+5) =1

------------ (x^2-9x+20)
(x^2-5x+5) -------------- =1

Answer 1

for an expression to equal 1 the exponent must equal 0.

so

x^2 -9x+20 = 0
(x-5)(x-4) = 0

x=5 or x=4

either way the bottom expression must equal 1 for the statement to be true, so:

x^2 - 5x + 5 = 1
x^2 -5x +4 = 0
(x-4)(x-1) =0
x=4 or x=1

Answer 2

Take the log of both sides. You may or may not know, the log of 1 is zero, so this will give you the following:

(x^2-9x+20)*log(x^2-5x+5 ) = 0

So, first, (x^2-9x+20) = 0, solve for x.

Second, log(x^2-5x+5 ) = 0

and you can put both sides back into an exponent, resulting in

x^2-5x+5 = 1, solve for x.

You should get anywhere between 2 and 4 values of x from these two quadratic equations.

To see why this is so, look back at the original question. For the left-hand side to equal one, either the exponent must be zero, for anything to the power of zero is one, or else the base must be one, because one to the power of anything is equal to one. Therefore we have the two quadratics to solve.

Answer 3

x^2 -9x + 20
x^2 - 5x - 4x + 20
x(x - 5) - 4(x - 5)
(x - 5)(x - 4)

x^2 - 5x + 5 = 1
x^2 -5x +4 = 0
x^2 - 4x - x + 4 = 0
x(x - 4) -1(x - 4) = 0
(x - 4)(x - 1) = 0
x = 4 or 1