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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

  • 4
    Can you solve the equation $ax=b(2x+1)$ for $x$?2010-10-20

4 Answers 4

4

Hint: The answer is using $\log_{10}$.

  • 0
    @giddy: Don't worry, it will become easier once you are more familiar.2010-10-20
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HINT$\ $ Putting $\rm\ a = 10^{\:b}\ $ yields $\rm\ x = 1/(b - 2)$

  • 2
    The point of the hint is to completely avoid logs!2010-10-20
1

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*}