1
$\begingroup$

I am having some trouble trying to find the single logarithm for the following:

$$\frac{1}{3} \ln(x+2)^3 + \frac{1}{2}[\ln x - \ln (x^2+3x+2)^2]$$

I understand that I have to use the addition and subtraction rules but the coefficients are confusing me. These are some of the steps I took:

$$\begin{align} \frac{1}{3} \ln(x+2)^3 + \frac{1}{2} [\ln \frac{x}{(x^2+3x+2)^2}] \end{align}$$

Now from here, if I used the addition rule then what would happen to the coefficients? and also, is there a way for me to simplify the powers $2$ and $3$?

Hints would be appreciated!

Thanks!

$\endgroup$

3 Answers 3

3
$\begingroup$

$$\frac{1}{3} \ln(x+2)^3 + \frac{1}{2}[\ln x - \ln (x^2+3x+2)^2]$$

$$=\ln(x+2)+\frac{1}{2} \ln(x)-\ln(x^2+3x+2)$$

$$=\ln\frac{(x+2)(\sqrt x)}{x^2+3x+2}$$

$$=\ln \frac{\sqrt{x}}{x+1}$$

Properties of $\ln(x)$ that have been used :

$$a \ln(b)= \ln (b^a), ln(a)+ln(b)=\ln(ab)$$

$\endgroup$
9
  • $\begingroup$ please see that this is not complete yet... $\endgroup$
    – user87543
    Commented Oct 10, 2013 at 11:47
  • $\begingroup$ Can you please explain where the power of two went? I understand that $\frac{1}{2}\ln x$ became $\ln \sqrt{x}$ but what about $\ln (x^2+3x+2)^2$? How did that just become $x^2+3x+2$? $\endgroup$
    – Jeel Shah
    Commented Oct 10, 2013 at 11:56
  • $\begingroup$ you do have $\frac{1}{2} \ln(x^2+3x+2)^2$... $\endgroup$
    – user87543
    Commented Oct 10, 2013 at 11:58
  • $\begingroup$ But the $\frac{1}{2}$ is not distributed is it? So how can it apply to both $\ln x$ and $\ln (x^2+3x+2)$? $\endgroup$
    – Jeel Shah
    Commented Oct 10, 2013 at 11:59
  • $\begingroup$ It is distributed.... You have $\frac{1}{2}[\ln x - \ln (x^2+3x+2)^2]$ Right???? $\endgroup$
    – user87543
    Commented Oct 10, 2013 at 12:01
2
$\begingroup$

Hints: $a\ln b=\ln b^a$, $\ln a + \ln b = \ln ab$, $x^2+3x+2=(x+1)(x+2)$.

$\endgroup$
0
$\begingroup$

Hint $$ b \cdot \log a = \log a^b$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .