What are the real solutions to the equation below?
(2x^2-5)
---------- ^ (x^2-3x) -1=0 (Note: ----- is a fracion bar and the fraction is being raised to (x^2-3x) and
3 and then you take the aswer to the fraction raised to (x^2-3x) and -1=0)
Also note that this is a quadratic...
Answer and Explination needed as soon as possible but will no be needed after tomorrow... Thank you!!!!
I read your equation as: [(2x^2 - 5)/3]^(x^2 - 3x) - 1 = 0, solve for x
3^(3 x - x^2) (2 x^2 - 5)^(x^2 - 3 x) - 1 = 0
Divide both sides by 3^(3 x - x^2):
(2 x^2 - 5)^(x^2 - 3 x) - 3^(x^2 - 3 x) = 0
Add 3^(x^2 - 3 x) to both sides:
(2 x^2 - 5)^(x^2 - 3 x) = 3^(x^2 - 3 x)
Take the natural logarithm of both sides and use the identity log(a^b) = b log(a):
log(2 x^2 - 5) (x^2 - 3 x) = log(3) (x^2 - 3 x)
(x^2 - 3 x) log(2 x^2 - 5) = x^2 log(2 x^2 - 5) - 3 x log(2 x^2 - 5):
x^2 log(2 x^2 - 5) - 3 x log(2 x^2 - 5) = log(3) (x^2 - 3 x)
Expand out terms of the right hand side:
x^2 log(2 x^2 - 5) - 3 x log(2 x^2 - 5) = log(3) x^2 - 3 log(3) x
Subtract x^2 log(3) - 3 x log(3) from both sides:
3 log(3) x - log(3) x^2 - 3 x log(2 x^2 - 5) + x^2 log(2 x^2 - 5) = 0
The left hand side factors into a product with four terms:
-x (x - 3) (log(3) - log(2 x^2 - 5)) = 0
Multiply both sides by -1:
x (x - 3) (log(3) - log(2 x^2 - 5)) = 0
Split x (x - 3) (log(3) - log(2 x^2 - 5)) into separate parts with additional assumptions.
Assume 2 x^2 - 5!=0 from log(2 x^2 - 5):
x - 3 = 0 for 2 x^2 - 5!=0
or x = 0 or log(3) - log(2 x^2 - 5) = 0
Add 3 to both sides:
x = 3 or x = 0 or log(3) - log(2 x^2 - 5) = 0
Subtract log(3) from both sides:
x = 3 or x = 0 or -log(2 x^2 - 5) = -log(3)
Multiply both sides by -1:
x = 3 or x = 0 or log(2 x^2 - 5) = log(3)
Cancel logarithms by taking exp of both sides:
x = 3 or x = 0 or 2 x^2 - 5 = 3
Add 5 to both sides:
x = 3 or x = 0 or 2 x^2 = 8
Divide both sides by 2:
x = 3 or x = 0 or x^2 = 4
Take the square root of both sides:
Answer: | x = 3 or x = 0 or x = 2 or x = -2