# Approximate values of these logs?

• Sep 17th 2009, 04:02 PM
Comical
Approximate values of these logs?
given that log_b 2=(approximately)0.756 and that log_b 10=(approximately)2.513, find the approximate values of each of ther following

a) log_b 40

for this one I got log_b 2^2x10. (x is times) then I got (.756)^2x10 =5.715. I have no idea if this is right? I am also unsure of..

b)log_b 1/2

I know I have to flip the fraction, so would that make it -log_b 2?

c) log_b 81/log_b 27

no idea what to do with this one?

d)log_b 1/b^3
same here. i just don't these.

there are a few more to this problem, but I got those ok. these have me stumped??
• Sep 17th 2009, 05:22 PM
Soroban
Hello, Comical!

Quote:

Given: .$\displaystyle \log_b(2)\:\approx\:0.756,\;\;\log_b(10)\:\approx\ : 2.513$
find the approximate value of each of the following:

$\displaystyle a)\;\;\log_b(40)$

$\displaystyle \log_b(40) \:=\:\log_b(2^2\cdot10) \:=\:\log_b(2^2) + \log_b(10) \:=\:2\log_b(2) + \log_b(10)$

Therefore: .$\displaystyle \log_b(40) \;=\;2(0.756) + 2.513 \;=\;4.025$

Quote:

$\displaystyle b)\;\;\log_b\!\left(\frac{1}{2}\right)$

$\displaystyle \log_b\!\left(\frac{1}{2}\right) \;=\;\log_b(2^{-1}) \;=\;-1\cdot \log_b(2) \;=\;-1(0.756) \;=\;-0.756$

Quote:

$\displaystyle c)\;\;\frac{\log_b(81)}{\log_b(27)}$
This one has nothing to do with the given decimals . . .

$\displaystyle \frac{\log_b(81)}{\log_b(27)} \;=\;\frac{\log_b(3^4)}{\log_b(3^3)} \;=\;\frac{4\cdot\log_b(3)}{3\cdot\log_b(3)} \;=\; \frac{4\cdot{\color{red}\rlap{//////}}\log_b(3)}{3\cdot{\color{red}\rlap{//////}}\log_b(3)} \;=\;\frac{4}{3}$

Quote:

$\displaystyle d)\;\;\log_b\!\left(\frac{1}{b^3}\right)$
$\displaystyle \log_b\!\left(\frac{1}{b^3}\right) \;=\;\log_b\left(b^{-3}\right) \;=\;-3\cdot\underbrace{\log_b(b)}_{\text{This is 1}} \;=\;-3\cdot1 \;=\;-3$

• Sep 17th 2009, 06:01 PM
Chinnie15
Woops, wrong account