# Math Help - Solving Logarithmic Fractions

1. ## Solving Logarithmic Fractions

I came upon this question to do with Logs and Fractions which I'm not too sure about. ${3log_{4} x} + \frac{\1}{2} = {5log_{8} x}$

I changed the bases to end up with $\frac{\3lnx}{ln4} - \frac{\5lnx}{ln8} = \frac{\1}{2}$

But forgot the rules about dividing logarithmic functions. Could somebody point me in the right direction please?

P.S. The question asks you to solve for x

2. ## Re: Solving Logarithmic Fractions

Take ln x out as a common factor from the left hand side. Divide the remaining term over then take both sides to the power of e.

3. ## Re: Solving Logarithmic Fractions

How do I take ln x out as a common factor? Since the its in the numerator. I don't quite understand that part

4. ## Re: Solving Logarithmic Fractions

Alternately. you could use the identity:

$\log_{b^n}a =\frac{\log_b(a)}{n}$ to write the equation as:

$3\log_{2^2}(x)+\frac{1}{2}=5\log_{2^3}(x)$

$3\frac{\log_2(x)}{2}+\frac{1}{2}=5\frac{\log_2(x)} {3}$

Now solve for $\log_2(x)$ and convert from logarithmic to exponential form...

5. ## Re: Solving Logarithmic Fractions

Thanks that solved it. I wasn't aware of the identity you mentioned but it makes this problem quite simple and I think its the way we are expected to solve it. Thanks!

6. ## Re: Solving Logarithmic Fractions

Continuing the method you began:

$\ln x\left(\frac{3}{2\ln 2}-\frac{5}{3\ln 2}\right)=-\frac{1}{2}$

$\ln x\left(-\frac{1}{6\ln 2}\right)=-\frac{1}{2}$

$\ln x=3\ln 2$

$\ln x=\ln 2^3$

$x=2^3$

Either method is probably what you are expected to know and use.