# Expressing In Terms of a And b

• Jan 15th 2009, 01:27 PM
Shapeshift
Expressing In Terms of a And b
If log (36) = a and log (125) = b

find log (1/12) in terms of a and b

I tried to do things like applying the product formula to log (36), making it log (12) + log (3), and then turning log (12) into -log (1/12), but could not get any further. Any help is appreciated.
• Jan 15th 2009, 01:38 PM
hmmmm
you would set k(log36/log125)=log(1/12) find k and the rewrite the log36 and log125 with a and b (im assuming log is the natural log in this case ln??)
• Jan 15th 2009, 03:16 PM
Shapeshift
I got...

log (1/12) = -a/2+b/3-1

could anyone confirm this? thanks!
• Jan 15th 2009, 07:14 PM
Soroban
Hello, Shapeshift!

Assuming that the logs are base-ten, you are right!

Quote:

If $\displaystyle \log(36) \,=\, a\:\text{ and }\:\log(125) \,=\, b$

find $\displaystyle \log\left(\tfrac{1}{12}\right)$ in terms of $\displaystyle a\text{ and }b.$

We have: .$\displaystyle \begin{array}{cccccccc} \log(36) \:=\:a & \Rightarrow & \log(6^2) \:=\:a & \Rightarrow & 2\log(6) \:=\:a & \Rightarrow & \log(6) \:=\:\frac{a}{2} \\ \\[-4mm] \log(125)\:=\:b & \Rightarrow & \log(5^3) \:=\:b & \Rightarrow & 3\log(5) \:=\:b & \Rightarrow & \log(5) \:=\:\frac{b}{3} \end{array}$

Therefore: .$\displaystyle \log\left(\frac{1}{12}\right) \;=\;\log\left(\frac{5}{6\cdot10}\right) \;=\;\log(5) - \log(6) - \log(10) \;=\;\boxed{\frac{b}{3} - \frac{a}{2} - 1}$