I think this is supposed to go here...anyways.. I need help with this log. Solve for x 5^(2x) - 5^(x) - 20 = 0 please and thank you...I've been stuck on it for a half hour now.
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Originally Posted by CenturionMonkey I think this is supposed to go here...anyways.. I need help with this log. Solve for x 5^(2x) - 5^(x) - 20 = 0 please and thank you...I've been stuck on it for a half hour now. Hello, use the substitution $\displaystyle y = 5^x$ and your equation becomes: $\displaystyle 5^{2x} - 5^x - 20 = 0~\longrightarrow~y^2-y-20=0$ This is a quadratic equation in y. Solve for y and don't forget to re-substitute to calculate x. I've got x = 1
ahh I see. Thanks a lot, yep the answer sheet says it's 1.
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