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So here is the problem:

Solved for a in terms of x:
$$a^{x} = 10^{2x + 1}$$

I tried:
$\displaystyle x \cdot \log(a) = (2x+1) \cdot \log\;10 $

$\displaystyle \frac{x}{2x + 1} = \frac{\log\;10} {\log\;a} $

But this is not going in the right direction, the answer according to the book is:
$$ \frac{1} {\log\;a - 2} $$

Excuse the 'power' tag for this question, there is no logarithm tag

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  • $\begingroup$ +1: For showing the effort you have put into the question. $\endgroup$ – Aryabhata Oct 20 '10 at 16:59
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    $\begingroup$ You are, in fact, going in the right direction. Keep going. $\endgroup$ – Qiaochu Yuan Oct 20 '10 at 17:00
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    $\begingroup$ Can you solve the equation $$ax=b(2x+1)$$ for $x$? $\endgroup$ – Mariano Suárez-Álvarez Oct 20 '10 at 17:01
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Hint: The answer is using $\log_{10}$.

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  • $\begingroup$ so I get to $ x = \frac{2x + 1}{log a} $ Please excuse my utter stupidity if I am unable to convert this into the answer! =S $\endgroup$ – gideon Oct 20 '10 at 17:06
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    $\begingroup$ @giddy: Can you solve $3x = 2x + 1$? What about $100x = 2x + 1$? $\endgroup$ – Aryabhata Oct 20 '10 at 17:07
  • $\begingroup$ (I can increase the teX font size) $\endgroup$ – gideon Oct 20 '10 at 17:09
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    $\begingroup$ @giddy: Don't you get log(a). x = 2x + 1? Is this not similar to 3x = 2x+1 or 100x = 2x +1? How did you solve those? $\endgroup$ – Aryabhata Oct 20 '10 at 17:29
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    $\begingroup$ @giddy, Don't divide yet! Get all the terms with an $x$ on the left hand side, and all the terms without an $x$ on the right hand side, and then you divide. $\endgroup$ – J. M. is a poor mathematician Oct 20 '10 at 17:30
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HINT$\ $ Putting $\rm\ a = 10^{\:b}\ $ yields $\rm\ x = 1/(b - 2)$

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  • $\begingroup$ you're using ln and e, not log base 10? Also, this makes me realize $$ x = \frac{2x + 1}{log a} $$ that is the answer.. so i could've left it at that since the question is only solve not show that this equals that? $\endgroup$ – gideon Oct 20 '10 at 17:35
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    $\begingroup$ The point of the hint is to completely avoid logs! $\endgroup$ – Bill Dubuque Oct 20 '10 at 17:52
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HINT:

Maybe you can find useful to look at Logarithm - Change of base, after solving your equation $\displaystyle \frac{x}{2x+1}=\frac{\text{log} 10}{\text{log}\thinspace a}$. You should finish with something like $x = \displaystyle \frac{1}{\frac{\displaystyle \text{log} \thinspace a}{\displaystyle \text{log} 10}-2}$

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Okay. So here is wat I got... \begin{align*} a^x &= 10^{2x}+1\\ x\log(a) &= 2x\log(10) + \log(10)\\ x\log(a) - 2x\log(10) &= \log (10)\\ x(\log(a)-2\log10) &= \log (10)\\ x &= (\log10) / (\log(a) - 2\log10)\\ x &= (1) / (\log(a) - 2)\\ \end{align*}

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