# Write as single logarithm

• Dec 17th 2008, 05:12 AM
mwok
Write 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 AM
iLikeMaths
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 AM
mwok
Ah. Is this correct?
• Dec 17th 2008, 06:02 AM
iLikeMaths
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 AM
mwok
Thanks!
• Dec 17th 2008, 06:04 AM
Moo
Quote:

Originally Posted by mwok
Ah. Is this correct?

There's just a problem with the $\displaystyle x^4-3^4$ part :D

$\displaystyle (x-3)^4=[(x-3)^2]^2=(x^2-6x+9)^2=$$\displaystyle (x^4+36x^2+81-12x^3-108x+18x^2)=(x^4-12x^3+54x^2-108x+81)$

But I think you can just keep it with $\displaystyle (x-3)^4$
• Dec 17th 2008, 06:15 AM
mwok
Wow.

Okay, so the final solution is attached.

EDIT:
Why does the "x" come out of ln(x)^2??
• Dec 17th 2008, 06:18 AM
iLikeMaths
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 AM
mwok
Quote:

Originally Posted by iLikeMaths
because of the two infront of it, note that this only works with logs/ln
$\displaystyle 2\ln(x) = \ln(x)^2$

....what happens to the "ln"?
• Dec 17th 2008, 06:20 AM
Moo
Quote:

Originally Posted by mwok
Wow.

Okay, so the final solution is attached.

EDIT:
Why does the "x" come out of ln(x)^2??

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 AM
Moo
Quote:

Originally Posted by iLikeMaths
because of the two infront of it, note that this only works with logs/ln
$\displaystyle 2\ln(x) = \ln(x)^2$

It's not a very correct writing.
Because one would understand it as "the square of the logarithm" whereas it is "the logarithm of the square"

So it should be written $\displaystyle \ln(x^2)$ (or $\displaystyle \ln x^2$, but it can still be ambiguous)
• Dec 17th 2008, 06:26 AM
mwok
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 AM
Moo
Quote:

Originally Posted by mwok
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?

We have :
$\displaystyle \ln((x-3)^4)+\ln(x^2)$

Now use the rule $\displaystyle \ln(a)+\ln(b)=\ln(ab)$ :

$\displaystyle =\ln[(x-3)^4 x^2]$
• Dec 17th 2008, 06:29 AM
mwok
Thanks!
• Dec 17th 2008, 06:33 AM
iLikeMaths
some times != means does not equal, well thats what i thought he/she meant