Question 1042180
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  9^{2x}\ +\ 4\,\cdot\,9^x\ =\ 21]


Let *[tex \Large u\ =\ 9^x], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  u^2\ +\ 4u\ -\ 21\ =\ 0]


Solve for *[tex \Large u]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (u\ -\ 3)(u\ +\ 7)\ =\ 0]


So *[tex \Large u\ =\ -7] or *[tex \Large u\ =\ 3]


Substitute back


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  9^x\ =\ -7]


Not possible.  Range of *[tex \Large y\ =\ a^x] is *[tex \Large \left\{y\ \in\ \mathbb{R}\ |\ y\ >\ 0\right\}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  9^x\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \log\left(9^x\right)\ =\ \log(3)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x\log(9)\ =\ \log(3)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x\ =\ \frac{\log(3)}{\log(9)}\ =\ \frac{\log(3)}{2\log(3)}\ =\ \frac{1}{2}]


Check


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  9^{2\cdot\frac{1}{2}}\ +\ 4\cdot9^{\frac{1}{2}}\ =\ 9\ +\ 4\,\cdot\,3\ =\ 9\ +\ 12\ = 21]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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