Question

HELP!!! 20 POINTS!!! URGENT!!!

Answer 1

We know that :

$\clubsuit$  $ln(A) - ln(B) = ln(\frac{A}{B})$

$\clubsuit$  $aln(x) = ln(x)^a$

$\clubsuit$  $ln(A) + ln(B) = ln(AB)$

Using above ideas we can solve the Problem :

⇒ $\frac{3}{8}ln(x + 3) = ln(x + 3)^\frac{3}{8}$

⇒ $ln(x - 3) - ln(x + 3)^\frac{3}{8} = ln[\frac{(x - 3)}{(x + 3)^\frac{3}{8}}]$

⇒ $4ln[\frac{(x - 3)}{(x + 3)^\frac{3}{8}}] = ln[\frac{(x - 3)}{(x + 3)^\frac{3}{8}}]^4 = ln[\frac{(x - 3)^4}{(x + 3)^\frac{3}{2}}]$

⇒ $\frac{1}{3}lnx + ln[\frac{(x - 3)^4}{(x + 3)^\frac{3}{2}}] = ln(x)^\frac{1}{3} + ln[\frac{(x - 3)^4}{(x + 3)^\frac{3}{2}}] = ln[\frac{\sqrt[3]{x}(x - 3)^4}{\sqrt{(x + 3)^{3}}}]$

Option 3 is the Answer