Express the given quantity as a single Logarithm $\frac{1}{5}\ln(x+2)^5+\frac{1}{2}[\ln x-\ln(x^2+3+2)^2)]$
Express the given quantity as a single Logarithm
$$\frac{1}{5}\ln(x+2)^5+\frac{1}{2}[\ln x-\ln(x^2+3x+2)^2)]$$
My answer is $\ln\dfrac{[(x+2)^5]^{\frac{1}{5}}\cdot x^2}{[(x^2+3x+2)^2]^{\frac{1}{2}}}$
What have I done wrong?
1 Answer
$\begingroup$Given $$\displaystyle \frac{1}{5}\ln(x+2)^5+\frac{1}{2}\left[\ln(x)-\ln(x^2+3x+2)^2\right]$$
$$\displaystyle \ln(x+2)^{\frac{5}{5}}+\ln(x)^{\frac{1}{2}}-\ln(x^2+3x+2)^{\frac{1}{2}} = \ln(x+2)+\ln (\sqrt{x})-\ln(x^2+3x+2)$$
So we get $$\displaystyle \ln\left[\frac{(x+2)\cdot \sqrt{x}}{x^2+3x+2}\right] = \ln\left[\frac{(x+2)\cdot \sqrt{x}}{(x+1)(x+2)}\right] = \ln\left[\frac{\sqrt{x}}{x+1}\right]$$
Above we have Used $$\bullet\; n\cdot \ln(m) = \ln(m)^{n}$$ and $$\displaystyle \bullet\; \ln(m)-\ln(n) = \ln\left(\frac{m}{n}\right)$$
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