1. ## 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:

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

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

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

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

$\displaystyle 9$

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

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

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

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

$\displaystyle 36$

2. Originally Posted by qbkr21
$\displaystyle 1.$ $\displaystyle ln(a)=2 , ln(b)=3 , ln(c)=10$
#1 should be:
$\displaystyle \frac{ln(a^{-2}b^1)}{ln(bc)^1}=$$\displaystyle \frac{-2ln(a) + ln(b)} {ln(b) + ln(c)}$

3. Hello, qbkr21!

You are misusing some log rules . . .

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

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

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

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

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

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

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

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

4. These answers are both off a bit... The answer to #1 should have been $\displaystyle \frac{-1}{8}$

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

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

7. ok thanks give me a moment... The latex software is still real new to me

8. Ok don't worry about the 1st one I figured it out and I know from now on thanks to Soroban that you treat $\displaystyle ln$ 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:

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

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