# Thread: Write as single logarithm

1. ## Write as single logarithm

Write (equation) as a single logarithm.

Attached is the equation and the solution.

Did I solve it correctly?

2. no you did not solve it correctly, remember the rules of log,

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

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

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

3. Ah. Is this correct?

4. 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 $4\ln(x-3) + 2\ln(x)$ with x equals 5 into your calculator you will get 5.999 or something

and if you put x=5 into the single logarithm $\ln(x-3)^4x^2$
you will get the same 5.99 so you know its correct

hope you understand

5. Thanks!

6. Originally Posted by mwok
Ah. Is this correct?
There's just a problem with the $x^4-3^4$ part

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

But I think you can just keep it with $(x-3)^4$

7. Wow.

Okay, so the final solution is attached.

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

8. because of the two infront of it, note that this only works with logs/ln
$2\ln(x) = \ln(x)^2$

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

....what happens to the "ln"?

10. 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 $\ln(x^2)$

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

What you wrote is :
$4 \log(x-3)=\log((x-3)^4)$

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

The final answer, $\ln((x-3)^4x^2)$ is correct.
But put as many parentheses as possible, because it's very easy to get wrong.

11. Originally Posted by iLikeMaths
because of the two infront of it, note that this only works with logs/ln
$2\ln(x) = \ln(x)^2$
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 $\ln(x^2)$ (or $\ln x^2$, but it can still be ambiguous)

12. 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?

13. 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 :
$\ln((x-3)^4)+\ln(x^2)$

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

$=\ln[(x-3)^4 x^2]$

14. Thanks!

15. some times != means does not equal, well thats what i thought he/she meant