3^y power because it was stated that y is greater than x so a number to the power of y will always be greater than a number to the power of x. Btw that looks a lot like the one I have lol
Step-by-step explanation:
o - odd number
e - even number
n + e =o, if n is odd...rule 1
n + e = e, if n is even...rule 2
n × o = o, if n is odd...rule 3
n × o = e, if n is even...rule 4
thus if 3n +6 = o,
then 3n must be odd, following rule 1
=> 3n = o, then n is an odd number, following rule 3
The expression into a single logarithm is ![log[(x)^{10}][(2)^{30}]](https://tex.z-dn.net/?f=log%5B%28x%29%5E%7B10%7D%5D%5B%282%29%5E%7B30%7D%5D)
Step-by-step explanation:
Let us revise some logarithmic rules
∵ 10 log(x) + 5 log(64)
- At first re-write 10 log(x)
∴ 10 log(x) = 
- Then re-write 5 log(64)
∴ 5 log(64) = 
∴ 10 log(x) + 5 log(64) =
+ 
- Use the 3rd rule above to make it single logarithm
∵
+
= ![log[(x)^{10}][(64)^{5}]](https://tex.z-dn.net/?f=log%5B%28x%29%5E%7B10%7D%5D%5B%2864%29%5E%7B5%7D%5D)
∴ 10 log(x) + 5 log(64) = ![log[(x)^{10}][(64)^{5}]](https://tex.z-dn.net/?f=log%5B%28x%29%5E%7B10%7D%5D%5B%2864%29%5E%7B5%7D%5D)
∵ 64 = 2 × 2 × 2 × 2 × 2 × 2
∴ We can write 64 as 
∴ 
- Multiply the two powers of 2
∴ 
∴ 10 log(x) + 5 log(64) = ![log[(x)^{10}][(2)^{30}]](https://tex.z-dn.net/?f=log%5B%28x%29%5E%7B10%7D%5D%5B%282%29%5E%7B30%7D%5D)
The expression into a single logarithm is ![log[(x)^{10}][(2)^{30}]](https://tex.z-dn.net/?f=log%5B%28x%29%5E%7B10%7D%5D%5B%282%29%5E%7B30%7D%5D)
Learn more:
You can learn more about the logarithmic functions in brainly.com/question/11921476
#LearnwithBrainly
The radius changes which is the constant when the circle decreases in size and remains at the same center point.
Answer:
3m^3/8 + 3
Step-by-step explanation:
It can't be solved but that's the most simplified version.