Andolph23 Messages 4 Reaction score 0 Thread starter Jul 30, 2011 #1 The Attempt at a Solution 2^x + 2^-x = 3 2^x + (1 / ((2^x)) = 3 (4^((x^2)) +1) / (2^x) = 3 (4^((x^2)) + 1) = 6^ x
The Attempt at a Solution 2^x + 2^-x = 3 2^x + (1 / ((2^x)) = 3 (4^((x^2)) +1) / (2^x) = 3 (4^((x^2)) + 1) = 6^ x
HallsofIvy Science Advisor Homework Helper Messages 42,895 Reaction score 983 Jul 30, 2011 #2 No, 2(3^x) is not the same 6^x.
gb7nash Homework Helper Messages 804 Reaction score 1 Jul 30, 2011 #3 Multiply both sides by 2x instead.
Andolph23 Messages 4 Reaction score 0 Jul 30, 2011 #4 My bad. I forgot that they must have the same base to multiply them. Also, If you're referring to the last step, I did multiply both sides by 2^x. I'm stuck at that point though.
My bad. I forgot that they must have the same base to multiply them. Also, If you're referring to the last step, I did multiply both sides by 2^x. I'm stuck at that point though.
GreenPrint Messages 1,186 Reaction score 0 Jul 30, 2011 #5 let u = 2^x solve the quadratic after solve for u solve for x
Andolph23 Messages 4 Reaction score 0 Jul 30, 2011 #6 Thanks that helped a lot. Thanks everyone for the quick replies