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
Answer:
h= 2/3/B -5
Step-by-step explanation:
move h=5 to the other side, divide b, then subtract 5
So the first thing you would have to is to find a common denominator so a common denominator between 11 and 15 is 165. 3/11 * 15 =45/165 next you would do 7/15 * 11=77/165 no that both denominators are the same we add the numerator but the denominator stays the same so 45/165 + 77/165 = 122/165
and since it can't be reduced any more 122/165 is your answer.
hope this helps :)
Answer:
4b^3 + 10b^2 - 32b + 15
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2(2b - 3)
Step-by-step explanation:
Take out the greatest common factor of (4b - 6) which is 2 and 3.
