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Please solve the following equation for $x$: $5^{2x}+(17/2)^x=9$

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What about using^{2x}%2B%2817%2F2%29^x%3D9+fo‌​r+x ? – Gottfried Helms Jan 24 '13 at 0:13

I don't see any nice way to do this. No closed form solution, certainly nothing pre-calculus. There is a (positive) real number $x$ that solves this, by continuity, but $x$ has no nice expression

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I know that x is equal to approximately 0.545411, I just don't know how to get to that solution... – Secret Secret Jan 24 '13 at 0:31
It can be inferred that the solution is unique (since $5^{2x}+(17/2)^x$ is increasing), and ranges between $0$ and $\ln_53$, but still no exact solution... – AndreasT Jan 24 '13 at 0:40

As Will Jagy already mentioned: there is no common closed form solution.
You can use newton-iteration (see wikipedia for more since you're tagging this with precalculus) to find an arbitrarily exact solution. In Pari/GP

f(x)  = 25^x+8.5^x-9       \\ define the function of which you search the zero
fd(x) = log(25)*25^x + log(8.5)*8.5^x   \\ and its derivative
\\ begin at a reasonable guess in the near of a solution
\\  sqrt(25)+sqrt(9) =8  and thus ~ 9 so x0 should be in the near of 1/2
x0 = 0.5   
\\ now iterate:
x0 = x0 - f(x0)/fd(x0)   \\ do this 7 times to arrive at dozens of correct digits
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