Logarithmic Equation

**fizz** · November 11th 2009, 05:23 PM

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

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

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

**Grandad** · November 11th 2009, 10:51 PM

Hello fizz

Originally Posted by fizz

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

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

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

$\log_{b} \frac{32}{5} = mr+ns$

$=m\log_{b} \frac{8}{45}+n\log_{b} \frac{135}{4}$

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

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

$\Rightarrow \frac{32}{5}=\frac{8^m135^n}{45^m4^n}$

Now express all these numbers in their prime factors:

$32=2^5,\quad 8=2^3,\quad 135 = 3^3\times5,\quad 45 = 3^2\times5,\quad 4=2^2$

$\Rightarrow \frac{2^5}{5^1}=2^5\,5^{-1}=\frac{2^{3m}\,3^{3n}\,5^n}{3^{2m}\,5^m\,2^{2n}} =2^{3m-2n}\,3^{3n-2m}\,5^{n-m}$

So if we now compare the indices, we get:

$2: 5 = 3m - 2n$

$5: -1 = n-m$

$\Rightarrow m=3, n=2$

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

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

Grandad

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