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April 22nd, 2014, 06:20 PM   #1
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Exponential Quadratic

This problem popped up on my homework:
$\displaystyle 7^{2x+1}-15•7^x+2=0$

Is the $\displaystyle 2x+1$ just a typo? Can this be done? Note that my teacher is very prone to making typos. I did the problem assuming that it was just $\displaystyle 7^{2x}$ and got $\displaystyle x=log_7\left({15\pm\sqrt{218}\over 2}\right)$. Thanks.
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April 22nd, 2014, 06:30 PM   #2
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\begin{align*}
7^{2x+1}-15 \cdot 7^x+2 &= 0 \\
7 \cdot 7^{2x}-15 \cdot 7^x+2 &= 0 \\
\left( 7 \cdot 7^x - 1\right) \left( 7^x - 2 \right) &= 0 \\
\left( 7^{x+1} - 1\right) \left( 7^x - 2 \right) &= 0 \\
\end{align*}

So $$\left(x + 1\right)\log{7} = 0 \; \Longrightarrow \; x = -1$$
or $$x\log{7} = \log{2} \; \Longrightarrow \; x = \frac{\log{2}}{\log{7}}$$

I like my numbers better than yours.
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Last edited by v8archie; April 22nd, 2014 at 06:36 PM.
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April 22nd, 2014, 06:53 PM   #3
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Quote:
Originally Posted by v8archie View Post
\begin{align*}
7^{2x+1}-15 \cdot 7^x+2 &= 0 \\
7 \cdot 7^{2x}-15 \cdot 7^x+2 &= 0 \\
\left( 7 \cdot 7^x - 1\right) \left( 7^x - 2 \right) &= 0 \\
\left( 7^{x+1} - 1\right) \left( 7^x - 2 \right) &= 0 \\
\end{align*}

So $$\left(x + 1\right)\log{7} = 0 \; \Longrightarrow \; x = -1$$
or $$x\log{7} = \log{2} \; \Longrightarrow \; x = \frac{\log{2}}{\log{7}}$$

I like my numbers better than yours.
I like yours a lot more too! Never thought of it like that. Thanks a lot
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April 22nd, 2014, 08:38 PM   #4
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An alternate approach: let y = 7^x and replace 7^(2x+1) with 7y^2.
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