3^[(sqrt x) -1]+3^[2-(sqrt x)]=12

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- Nov 20th 2012, 11:54 AMDarkAngleHelp me solve this equation please.
3^[(sqrt x) -1]+3^[2-(sqrt x)]=12

- Nov 20th 2012, 12:08 PMMarkFLRe: Help me solve this equation please.
This forum is for introductions. The forums most appropriate for this topic is either algebra or pre-calculus.

Anyway, I think I would rewrite the equation as follows:

$\displaystyle 3^{\sqrt{x}-1}+3^{2-\sqrt{x}}-12=0$

Multiply through by $\displaystyle 3^{\sqrt{x}+1}=3\cdot3^{\sqrt{x}}$:

$\displaystyle \left(3^{\sqrt{x}} \right)^2-36\left(3^{\sqrt{x}} \right)+27=0$

Now you have a quadratic in $\displaystyle 3^{\sqrt{x}$. - Nov 20th 2012, 12:14 PMpickslidesRe: Help me solve this equation please.
$\displaystyle \displaystyle a^{m-n} = \frac{a^m}{a^n}$

Does this make life easier? - Nov 20th 2012, 01:28 PMDarkAngleRe: Help me solve this equation please.
[3^(sqrtx)]/3 + 3^2/3(sqrtx)=12?

- Nov 20th 2012, 01:55 PMskeeterRe: Help me solve this equation please.
some advice ... look at post #2 again