Write (equation) as a single logarithm.
Attached is the equation and the solution.
Did I solve it correctly?
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
$\displaystyle \ln(x-3)^4x^2$
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
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.