Theses are just some problems i did not understand how to do so if you could please explain and write the answer(:

Question

anything helps
thank you(:

2. Simplify
---(3x^11z^5)^3
b. (3x^5)^ -3 / 2x-5(3y)^4
c.3x^5)^ -3 / 2x ^ -5(3y)^4 (same as b but - 5 is the exponent)

3a.x^2 / 5 + x-5 / 6=0 (proportion, and then i got to the quadratic equation and after i did that i got 5/3 and -2.5 correct?)
3b.18=2(3)^x+3 (x+3 are exponents)
c. 128=4^3x (whats x)

Answer 1

2.
(3x^11z^5)^3 = 27 x^33 z^15

b) and c) are not clear since you don't use proper brackets

3a) again, your lack of brackets make it ambiguous

I will assume you mean
x^2/5 + (x-5)/6 = 0
then
x^2/5 = (5-x)/6
cross-multiply
6x^2 = 25 - 5x
6x^2 + 5x - 25 = 0
your answer is correct

3b)
18 = 2(3)^(x+3)
9 = 3^(x+3)
3^2 = 3^(x+2)
so x+2 = 2
x = 0

c)
128 = 4^(3x)
2^7 = (2^2)^(3x)
2^7 = 2^(6x)
6x = 7
x = 7/6

Answer 2

this is how 2b and 2c is written:
(3x^5)^-3
-----------
2x^-5(3y)^4

Does that help?

(3x^5)^-3
-----------
2x-5(3y)^4

Answer 3

that is what think alyssa is trying to say?