## Solving for "X"-ponents

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
GreenLantern
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Joined: Sat Mar 07, 2009 10:47 pm
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### Solving for "X"-ponents

I have a relatively simple equation, but am not sure I'm going about it in the right way.

$2^{2x+1}=9$

My assumption is that I need to get the bases equal. The only way I know how to do this is to raise them to equalizing powers, the only number that can do that is 0. (This is the part I'm unsure of.)

$2^{2x+1}=9 \Rightarrow 2^{2x+1+0}=9^{1+0}$

After that the problem is really simple. The bases can be nixed and equations solved.

$2^{2x+1+0}=9^{1+0} \Rightarrow$
$x=\frac{0}{2}\$

But that makes no sense... Plugging that in gets an obviously wrong answer, so I return to the part I'm unsure of and am now here.

stapel_eliz
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Joined: Mon Dec 08, 2008 4:22 pm
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GreenLantern wrote:$2^{2x+1}=9$

My assumption is that I need to get the bases equal.

Since 9 is not a whole-number or fractional power of 2, no, trying to get equal bases (so you can equate the powers) isn't going to work. You're stuck with logarithms:

. . . . .$\ln\left(2^{2x+1}\right)\, =\, \ln(9)$

. . . . .$(2x\, +\, 1)\ln(2)\, =\, \ln(9)$

Divide through by $\ln(2)$, subtract the $1$, and then divide through by the $2$ to get an expression for $x$.

Don't simplify (that is, plug it into your calculator to find a decimal approximation) unless instructed to do so.