We know that :



Using above ideas we can solve the Problem :
⇒ 
⇒ ![ln(x - 3) - ln(x + 3)^\frac{3}{8} = ln[\frac{(x - 3)}{(x + 3)^\frac{3}{8}}]](https://tex.z-dn.net/?f=ln%28x%20-%203%29%20-%20ln%28x%20%2B%203%29%5E%5Cfrac%7B3%7D%7B8%7D%20%3D%20ln%5B%5Cfrac%7B%28x%20-%203%29%7D%7B%28x%20%2B%203%29%5E%5Cfrac%7B3%7D%7B8%7D%7D%5D)
⇒ ![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}}]](https://tex.z-dn.net/?f=4ln%5B%5Cfrac%7B%28x%20-%203%29%7D%7B%28x%20%2B%203%29%5E%5Cfrac%7B3%7D%7B8%7D%7D%5D%20%3D%20ln%5B%5Cfrac%7B%28x%20-%203%29%7D%7B%28x%20%2B%203%29%5E%5Cfrac%7B3%7D%7B8%7D%7D%5D%5E4%20%3D%20ln%5B%5Cfrac%7B%28x%20-%203%29%5E4%7D%7B%28x%20%2B%203%29%5E%5Cfrac%7B3%7D%7B2%7D%7D%5D)
⇒ ![\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}}}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7Dlnx%20%2B%20ln%5B%5Cfrac%7B%28x%20-%203%29%5E4%7D%7B%28x%20%2B%203%29%5E%5Cfrac%7B3%7D%7B2%7D%7D%5D%20%3D%20ln%28x%29%5E%5Cfrac%7B1%7D%7B3%7D%20%2B%20ln%5B%5Cfrac%7B%28x%20-%203%29%5E4%7D%7B%28x%20%2B%203%29%5E%5Cfrac%7B3%7D%7B2%7D%7D%5D%20%3D%20ln%5B%5Cfrac%7B%5Csqrt%5B3%5D%7Bx%7D%28x%20-%203%29%5E4%7D%7B%5Csqrt%7B%28x%20%2B%203%29%5E%7B3%7D%7D%7D%5D)
Option 3 is the Answer
<span>The congruent complements theorem 2 <span> angles are complements of the same angle (or of congruent angles), then the two angles are congruent.</span></span>
to find the radius you divide the circumference by 2pi
area is pi times radius to the 2nd power
APR is different than other compounding periods because you would need to find some equivelancy to compare things at.