TITLE: Stuck on an 'advanced logarithm problem': $2 \log_2 x - \log_2 (x - \tfrac1 2) = \log_3 3$
QUESTION [6 upvotes]: 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?

REPLY [3 votes]: 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!