Answer:
5 and -5
Step-by-step explanation:
Replace y with 0 and solve for x
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:
The zeroes of this function are x=-5 and x=4
X=4
This is because
(4)3-2=10
12-2=10
10=10
This makes the statement true!
Hope this helps have a good day!
Here you go. Do you think that you could help me with my question
5a – 10b = 45
5a – 10(3) = 45
5a – 30 = 45
+30 +30
5a = 75/5
<span>a = 15</span>