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I'm stuck on solving what my textbook calls an "advanced logarithm problem". Basically, it's a logarithmic equation with logarithms of different bases on either side. My exercise looks like this:

$$2 \log_2 x - \log_2 (x - \tfrac1 2) = \log_3 3$$

To start off, I used the power rule to simplify the first term to get this:

$$\log_2 x^2 - \log_2 (x - \tfrac1 2) = \log_3 3$$

Then I used the quotient rule to get this:

$$\frac {x^2} {x - \frac 1 2} = \log_3 3$$

Then I turned the logarithmic equation into an exponential equation to get this:

$$3^{\frac {x^2} {x - \frac 1 2}} = 3$$

Now, however, I'm unsure of how to proceed. The textbook has neither explained to me how to simplify such complex exponents nor do such exponents have any precedence. I'm therefore assuming that I went wrong somewhere previously in solving the problem, but as far as I can tell I did everything by the book.

Did I go wrong? And if not, am I really supposed to simplify that exponent?

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    $\begingroup$ You lost a logarithm when applying the quotient rule. $\endgroup$ Jul 3, 2013 at 1:44
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    $\begingroup$ 1.First turn everything to the same base, use $\log_b a = \frac {\ln a}{\ln b} $ $\endgroup$
    – jimjim
    Jul 3, 2013 at 1:45
  • $\begingroup$ @Daniel Oops, where? I don't see it... -edit- Got it. $\endgroup$ Jul 3, 2013 at 1:46
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    $\begingroup$ $\log_2 a - \log_2 b = \log_2 (a/b)$. You should get $\log_2 \frac{x^2}{x-1/2} = \log_3 3$. $\endgroup$ Jul 3, 2013 at 1:49
  • $\begingroup$ Ah, okay, thanks Daniel. That makes sense. :) Slowly figuring this out. $\endgroup$ Jul 3, 2013 at 1:51

2 Answers 2

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Sorry, read too quickly at first. You were on the right track - but you lost a logarithm at this step: $$\frac{x^2}{x-\frac{1}{2}}=\log_3(3)$$ It should be $$\log_2\left(\frac{x^2}{x-\frac{1}{2}}\right)=\log_3(3)$$ Now, what is $\log_3(3)$?

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  • $\begingroup$ I'm very bad at logarithms. My work with them has been half-guesswork thus far. 1? $\endgroup$ Jul 3, 2013 at 1:49
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    $\begingroup$ That's exactly right - the exponent that you have to raise $3$ to in order to obtain $3$ is just $1$, since $3^1=3$. In the same way, since $3^2=9$, we have $\log_3(9)=2$. $\endgroup$ Jul 3, 2013 at 1:50
  • $\begingroup$ Awesome, thanks. I'm not writing out my steps in the comment but I'm working through the rest of the problem on paper right now. I think I'm on to the solution. $\endgroup$ Jul 3, 2013 at 1:54
  • $\begingroup$ It was losing the logarithm that did it--I've arrived at the answer $x = 1$. Thanks for helping me work through this! Plus you helped clarify my thinking on logarithms which is awesome. Accepted. $\endgroup$ Jul 3, 2013 at 2:00
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Zev Chonoles pointed out your mistake, so continuing from that:

From the definition of the logarithm, $\log_3 3=1$ (think $3$ to the what power equals $3$?)

So $\log_2\left(\frac{x^2}{x-\frac{1}{2}}\right)=1$

Again if we apply the definition of the logarithm,

$2^1=\left(\frac{x^2}{x-\frac{1}{2}}\right)$ , or simply just

$2=\left(\frac{x^2}{x-\frac{1}{2}}\right)$

Multiplying both sides by $x-\frac 12$ ,

$2x-1=x^2$

$x^2-2x+1=0$

From here you can use any method you would like to solve this quadratic, but I think the fastest is to notice that this factors out to

$(x-1)^2=0$

So $x=1$

Hope this helps!

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  • $\begingroup$ I appreciate the help, but the reason Zev didn't finish the problem for me was that I marked it as homework and would benefit more from figuring it out by myself. Thanks for pointing out the solution, though! :) $\endgroup$ Jul 3, 2013 at 1:57
  • $\begingroup$ No problem, and sorry my mistake I forgot it was homework. $\endgroup$
    – Ovi
    Jul 3, 2013 at 1:58

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