# LOgarithm proBLeM

• Nov 29th 2008, 05:07 AM
mathbuoy
LOgarithm proBLeM
If log[a^2]b^2 - log[b]a = 4, find log[a]b.

#characters in [ ] is at lower 'case' , attached to 'log'.

*show working and explanation, pls...(Doh)
• Nov 29th 2008, 05:30 AM
running-gag
Use the formula
$\displaystyle log_a(x)\;=\;\frac{ln(x)}{ln(a)}$
• Nov 29th 2008, 02:28 PM
Soroban
Hello, mathbuoy!

This is a strange one . . . I don't recommend natural logs.

Quote:

If $\displaystyle \log_{a^2}(b^2) - \log_b(a) \:=\: 4$, find $\displaystyle \log_a(b)$
This is what I do when I encounter a strange base . . .

Let $\displaystyle \log_{a^2}(b^2) \:=\:p \quad\Rightarrow\quad(a^2)^p = b^2 \quad\Rightarrow\quad a^{2p} = b^2$

Take logs (base $\displaystyle a$): .$\displaystyle \log_a(a^{2p}) \:=\:\log_a(b^2) \quad\Rightarrow\quad 2p\log_a(a) \:=\:2\log_a(b) \quad\Rightarrow\quad p \:=\:\log_a(b)$

. . Hence: .$\displaystyle \log_{a^2}(b^2) \;=\;\log_a(b)$

And equation becomes: .$\displaystyle \log_a(b) - \log_b(a) \;=\;4$

Use this identity on the second term: .$\displaystyle \log_b(a) \:=\:\frac{1}{\log_a(b)}$
. . and we have: .$\displaystyle \log_a(b) - \frac{1}{\log_a(b)} \;=\;4$

Multiply by $\displaystyle \log_a(b)\!:\quad \left[\log_a(b)\right]^2 - 1 \;=\;4\log_a(b) \quad\Rightarrow\quad\left[\log_a(b)\right]^2 - 4\log_a(b) - 1 \;=\;0$

Quadratic Formula: .$\displaystyle \log_a(b) \;=\;\frac{4 \pm\sqrt{20}}{2}$

. . Therefore: .$\displaystyle \log_a(b)\;=\;2 \pm\sqrt{5}$

• Nov 30th 2008, 04:50 AM
running-gag
Quote:

Originally Posted by Soroban
Hello, mathbuoy!

This is a strange one . . . I don't recommend natural logs.

It works well anyway !
We have to find $\displaystyle X\;=\;log_a(b)\;=\;\frac{ln(b)}{ln(a)}$

$\displaystyle log_{a^2}(b^2)\;=\;\frac{ln(b^2)}{ln(a^2)}$

$\displaystyle log_{a^2}(b^2)\;=\;\frac{2ln(b)}{2ln(a)}$

$\displaystyle log_{a^2}(b^2)\;=\;\frac{ln(b)}{ln(a)}$

$\displaystyle log_{a^2}(b^2)\;=\;X$

Equation becomes
X - 1/X = 4
X² - 4X - 1 = 0