Express the quantity as a single logarithm: $ \frac 13 \ln (x+2)^3 + \frac 12[ \ln x - \ln (x^2+3x+ 2)^2]$

Question

Express the quantity as a single logarithm: $ \frac 13 \ln (x+2)^3 + \frac 12[ \ln x - \ln (x^2+3x+ 2)^2]$

The answer in the book is ln $\frac {\sqrt{x}}{x+1}$

If am not allowed to to cancel terms when they are sums, how was this answered calculated?

Answer 1

for $x>0$ we get $$\log(x+2)+\log(x^{1/2})-\log(x^2+3x+2)=\log((x+2)x^{1/2})-\log(x^2+3x+2)$$ and finally we get $$\log\left(\frac{(x+2)x^{1/2}}{x^2+3x+2}\right)$$

Answer 2

You can use the rule $\ln A + \ln B = \ln (AB)$.

Answer 3

Your mistake lies in the division. It's true that: $$-\log x = \log x^{-1}=\log\left(\frac{1}{x}\right)$$ It is not however true that: $$-\log x = \frac{1}{\log x}$$

In other words, your third line should read: $$\ln (x+2) + \ln \sqrt{x} - \ln \frac{1}{x^2+3x+2}$$ $$=\ln \left(\frac{\sqrt{x}(x+2)}{x^2+3x+2}\right)$$ Can you continue from there?