SOLUTION: Hi, this is review in my Calculus book... Solve: log(base x) (2x-3)=2 for x. I tried doing this: 2x-3=x^2 x^2-2x+3=0 (x-3)(x+1)=0 So x = 3 and x = 1. But then I thought it mi

Question 497909: Hi, this is review in my Calculus book... Solve: log(base x) (2x-3)=2 for x.
I tried doing this:
2x-3=x^2
x^2-2x+3=0
(x-3)(x+1)=0
So x = 3 and x = 1.
But then I thought it might be False since if you put 3 and 1 into the first eqation that it wouldn't come out to 2.
Thanks for helping
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Solve: log(base x) (2x-3)=2 for x
Base(x) raised to log of number(2)=number(2x-3)
2x-3=x^2
x^2-2x+3=0
(x-3)(x+1)=0
So x = 3 and x = 1
**
You got the basic log quadratic equation correct, but you goofed up on its solution.
a=1, b=-2, c=3
The discriminant, √(b^2-4ac)=√(4-4*1*3)=√(4-12)=√-8
There is no solution, because x is not a real number
..
Good Luck on your calculus class!
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