Question

THE LAST PART TO THE QUESTION ONLYPLEASE DO NPT ANSWER THE FIRST 2 PARTS ONLY THE LAST POART

Answer 1

#12 c

The given expression is

$\ln (\frac{e^{6x}(x^2+14)^{\frac{1}{9}}}{(x-4)^4})$

We will use the rule of expansion of ln

$\ln (\frac{ab}{c})=la(a)+\ln (b)-\ln (c)$$\ln (e^{6x})+\ln (x^2+14)^{\frac{1}{9}}-\ln (x-4)^4$

Use the rule

$\begin{gathered} \ln (a)^n=n\ln (a) \\ \ln (e^n)=n \end{gathered}$$6x+\frac{1}{9}\ln (x^2+14)-4\ln (x-4)$

The answer is

$\ln (\frac{e^{6x}(x^2+14)^{\frac{1}{9}}}{(x-4)^4})=6x+\frac{1}{9}\ln (x^2+14)-4\ln (x-4)$