# Natural Log Seperation

• Jan 29th 2007, 01:12 PM
qbkr21
Natural Log Seperation
I worked these math problems out, however the system says they are incorrect. I would appreciate it if you could please check over my work, and maybe even tell me where I went wrong:

$1.$ $ln(a)=2 , ln(b)=3 , ln(c)=10$

$\frac{ln(a^{-2}b^1)}{ln(bc)^1}$

$-2ln(a)+1ln(b)-1ln(b)-1ln(c)$

$-2(2)+1(2)-1(3)-1(5)$

$9$

$2.$ $ln(a)=2 , ln(b)=3 , ln(c)=10$

$ln(c^{-4})*ln(\frac{a}{b^{-4}})^4$

$-4ln(c)+4ln(a)+16ln(b)$

$-4(5)+4(2)+16(3)$

$36$
• Jan 29th 2007, 03:11 PM
Plato
Quote:

Originally Posted by qbkr21
$1.$ $ln(a)=2 , ln(b)=3 , ln(c)=10$

#1 should be:
$\frac{ln(a^{-2}b^1)}{ln(bc)^1}=$ $\frac{-2ln(a) + ln(b)} {ln(b) + ln(c)}$
• Jan 29th 2007, 03:11 PM
Soroban
Hello, qbkr21!

You are misusing some log rules . . .

Quote:

$1.\;\ln(a)=2,\;\ln(b)=3,\;\ln(c)=10$

$\frac{\ln(a^{-2}b^1)}{\ln(bc)^1} \;= \;-2\ln(a)+1\ln(b)-1\ln(b)-1\ln(c)$ . . . no

$\frac{\ln(a^{-2}b)}{\ln(bc)} \:=\:\frac{-2\ln(a) + \ln(b)}{\ln(b) + \ln(c)} \:=\:\frac{-2(2) + 3}{3 + 10} \:=\:-\frac{1}{13}$

Quote:

$2.\;\ln(a)=2,\;\ln(b)=3,\;\ln(c)=10$

$\ln(c^{-4})\cdot\ln(\frac{a}{b^{-4}})^4 \;= \;-4\ln(c) + 4\ln(a) + 16\ln(b)$ . . . no

$\ln(c^{-4})\cdot\ln\left(\frac{a}{b^{-4}}\right)^4\:=\:-4\ln(c)\cdot 4\ln\left(\frac{a}{b^{-4}}\right)$

. . $=\:-16\ln(c)\left[\ln(a) - \ln(b^{-4})\right] \:= \:-16\ln(c)\left[\ln(a) + 4\ln(b)\right]$

. . $=\:-16(10)\left[2 + 4\cdot3\right] \;=\;-160(14)\:=\:-2240$

• Jan 29th 2007, 03:56 PM
qbkr21
These answers are both off a bit... The answer to #1 should have been $\frac{-1}{8}$
• Jan 29th 2007, 04:01 PM
topsquark
Quote:

Originally Posted by qbkr21
These answers are both off a bit... The answer to #1 should have been $\frac{-1}{8}$

Their answers are correct. I not only derived their expressions, but I plugged the original problem into my calculator and verified their results. Could there be a typo in your problem? That "1" in the denominator is in a strange place.

-Dan
• Jan 29th 2007, 04:04 PM
Plato
Quote:

Originally Posted by qbkr21
These answers are both off a bit... The answer to #1 should have been $\frac{-1}{8}$

If that is true, then you have given the problem incorrectly.
Please review what you have written.
It is also possible that the text is wrong!
• Jan 29th 2007, 04:12 PM
qbkr21
ok thanks give me a moment... The latex software is still real new to me
• Jan 29th 2007, 04:23 PM
qbkr21
Ok don't worry about the 1st one I figured it out and I know from now on thanks to Soroban that you treat $ln$ on top and bottom as two separate pieces. Problem #2 is a follows:

I think that the values might be off, use these values instead:

$\ ln(a)=2 \ ln(b)=3 \ ln(c)=5$

$(\ln\ c^{-4})(\ln \frac{a}{b^{-4}})^{4}$