# Thread: Logarithmic Equation

1. ## Logarithmic Equation

Let r = $\displaystyle \log_{b} \frac{8}{45}$ and s = $\displaystyle \log_{b} \frac{135}{4}$.

Find an ordered pair of integers (m,n) such that $\displaystyle \log_{b} \frac{32}{5} = mr+ns$.

Can someone help me set this up? I don't see how to solve for 2 variables with only one equation.

2. Hello fizz
Originally Posted by fizz
Let r = $\displaystyle \log_{b} \frac{8}{45}$ and s = $\displaystyle \log_{b} \frac{135}{4}$.

Find an ordered pair of integers (m,n) such that $\displaystyle \log_{b} \frac{32}{5} = mr+ns$.

Can someone help me set this up? I don't see how to solve for 2 variables with only one equation.
$\displaystyle \log_{b} \frac{32}{5} = mr+ns$
$\displaystyle =m\log_{b} \frac{8}{45}+n\log_{b} \frac{135}{4}$

$\displaystyle =\log_{b}\left( \frac{8}{45}\right)^m+\log_{b}\left( \frac{135}{4}\right)^n$

$\displaystyle =\log_{b}\left( \frac{8}{45}\right)^m\left( \frac{135}{4}\right)^n$
$\displaystyle \Rightarrow \frac{32}{5}=\frac{8^m135^n}{45^m4^n}$

Now express all these numbers in their prime factors:
$\displaystyle 32=2^5,\quad 8=2^3,\quad 135 = 3^3\times5,\quad 45 = 3^2\times5,\quad 4=2^2$
$\displaystyle \Rightarrow \frac{2^5}{5^1}=2^5\,5^{-1}=\frac{2^{3m}\,3^{3n}\,5^n}{3^{2m}\,5^m\,2^{2n}} =2^{3m-2n}\,3^{3n-2m}\,5^{n-m}$

So if we now compare the indices, we get:

$\displaystyle 2: 5 = 3m - 2n$

$\displaystyle 5: -1 = n-m$

$\displaystyle \Rightarrow m=3, n=2$

Check the power of $\displaystyle 3: 0=3n-2m$, which is correct.

So the ordered pair is $\displaystyle (3,2)$.

Grandad