2^x-2^-x/3=4

jibbajbr · Feb 14, 2013, 06:22 AM

I need to solve for x and have no clue what I'm doing

ebaines · Feb 15, 2013, 07:12 AM

Please clarify- do you mean:

$<br /> 2^x - 2^{(\frac {-x}{3})} = 4$

or:

$<br /> 2^x - \frac {2^{(-x)}} 3 = 4<br />$

?

Solving the first involves solving a 4th order polynomial: let $u = 2^{x/3}$ and you get $u^4-4u-1 = 0$ . Solve for u, then

$x = 3 \frac {\ln(u)}{\ln(2)}$

Solving the second is easier. Let $u = 2^x$ , which leads to $u^2-4u - 1/3 = 0$ , which can be solved with the quadratic formula. Then $x = \frac { \ln(u)}{ \ln(2)}$

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