Solve for X AlgebraicallyDate: 07/24/2003 at 21:09:34 From: John Subject: Exponents Many people (including myself) have been stumped by this question: Solve for x algebraically, (no graphing) 2^x + 2^(-x) = 5 I tried doing different things, but ended up just going in a circle. Date: 07/24/2003 at 23:59:07 From: Doctor Ian Subject: Re: Exponents Hi John, Let's begin by writing it this way: 1 2^x + --- = 5 2^x Now we have a fraction, and the easiest way to get rid of a fraction is to multiply both sides by the denominator: (2^x)(2^x) + 1 = 5(2^x) (2^x)^2 + 1 = 5*(2^x) Hmmmm. This looks a little like a quadratic equation, doesn't it? Suppose we make the substitution u = 2^x Then we have u^2 + 1 = 5u u^2 - 5u + 1 = 0 Now, suppose we solve this and end up with u = [something] This means that 2^x = [something] which is simpler to deal with. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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