Write (equation) as a single logarithm.

Attached is the equation and the solution.

Did I solve it correctly?

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- Dec 17th 2008, 05:12 AMmwokWrite as single logarithm
Write (equation) as a single logarithm.

Attached is the equation and the solution.

Did I solve it correctly? - Dec 17th 2008, 05:54 AMiLikeMaths
no you did not solve it correctly, remember the rules of log,

$\displaystyle 4\ln(x-3)$ is the same as $\displaystyle \ln(x-3)^4$

and $\displaystyle 2\ln(x)$ is the same as $\displaystyle \ln(x)^2$

also

$\displaystyle \ln(x-3)^4$ + $\displaystyle \ln(x)^2$ is the same as

$\displaystyle \ln(x-3)^4x^2$ - Dec 17th 2008, 05:57 AMmwok
Ah. Is this correct?

- Dec 17th 2008, 06:02 AMiLikeMaths
and there it is, as a single logarithm, u can try to put any number back into the equation to check your working e.g. let x = 5

if u put this $\displaystyle 4\ln(x-3) + 2\ln(x)$ with x equals 5 into your calculator you will get 5.999 or something

and if you put x=5 into the single logarithm $\displaystyle \ln(x-3)^4x^2$

you will get the same 5.99 so you know its correct

hope you understand - Dec 17th 2008, 06:04 AMmwok
Thanks!

- Dec 17th 2008, 06:04 AMMoo
- Dec 17th 2008, 06:15 AMmwok
Wow.

Okay, so the final solution is attached.

EDIT:

Why does the "x" come out of ln(x)^2?? - Dec 17th 2008, 06:18 AMiLikeMaths
because of the two infront of it, note that this only works with logs/ln

$\displaystyle 2\ln(x) = \ln(x)^2$ - Dec 17th 2008, 06:20 AMmwok
- Dec 17th 2008, 06:20 AMMoo
No, it's $\displaystyle \ln(x^2)$

It's not really $\displaystyle 4 \log(x-3) !=\log(x^4-3^4)$. I mean... stated this way, it looks like you didn't understand your mistake.

What you wrote is :

$\displaystyle 4 \log(x-3)=\log((x-3)^4)$

But $\displaystyle (x-3)^4 !=x^4-3^4$, this was your mistake.

The final answer, $\displaystyle \ln((x-3)^4x^2)$ is correct.

But put as many parentheses as possible, because it's very easy to get wrong. - Dec 17th 2008, 06:21 AMMoo
- Dec 17th 2008, 06:26 AMmwok
Ah I see how the exponents work now BUT why does x^2 come out of the ln(x^2)? What happens to the ln?

- Dec 17th 2008, 06:28 AMMoo
- Dec 17th 2008, 06:29 AMmwok
Thanks!

- Dec 17th 2008, 06:33 AMiLikeMaths
some times != means does not equal, well thats what i thought he/she meant