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$\begingroup$

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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    +1: For showing the effort you have put into the question.2010-10-20
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    You are, in fact, going in the right direction. Keep going.2010-10-20
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    Can you solve the equation $$ax=b(2x+1)$$ for $x$?2010-10-20

4 Answers 4

4

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

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    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! =S2010-10-20
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    @giddy: Can you solve $3x = 2x + 1$? What about $100x = 2x + 1$?2010-10-20
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    (I can increase the teX font size)2010-10-20
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    @giddy: I have edited the answer to increase the font size. You can either use `\displaystyle` or use two dollar signs.2010-10-20
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    @Moron so first equation $$ (3x-2x=1) x = 1 $$ and the second $$ x = \frac {1}{98} $$ Correct? Dont see how it related to simplifying this==> $$ x = \frac {2x + 1}{log a} $$2010-10-20
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    Just after taking log what did you get?2010-10-20
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    in which equation? my problem? Thats what I get right, up here?2010-10-20
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    @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?2010-10-20
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    @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.2010-10-20
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    ...and a final hint: $ax-bx=(a-b)x$2010-10-20
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    GOT IT... ugh feel like really kicking myself in the arse... log(a). x = 2x + 1 ==> 2x - xloga = 1 ==> x(2 - log a) = 1 >> x = 1/2-log a2010-10-20
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    thank you for the annoying hints making me FIND my own answer... I know it took more **time** ...I wish I could see through problems like this but its mostly my frustration and anxiety i think thats getting in the way!!2010-10-20
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    oopsy... i made a sign mistake in the previous comment!2010-10-20
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    @giddy: Don't worry, it will become easier once you are more familiar.2010-10-20
1

HINT$\ $ Putting $\rm\ a = 10^{\:b}\ $ yields $\rm\ x = 1/(b - 2)$

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    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?2010-10-20
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    The point of the hint is to completely avoid logs!2010-10-20
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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*}