Question

Please Help Me Please

Answer 1

@Directrix

Answer 2

ok.. the key to this is knowing about powers of 2

\[\frac{1}{2} = 2^{-1} ~~~~and~~~~~32 = 2^5\]

so the problem now becomes

\[(2^{-1})^{2x} = 2^5\]

if you know the index law for power of a power you should be able to find a solution

Answer 3

immm 2 -1 x 2x gives you 2x so 2^ 2x =2^5. I don't know how to solve this are you suppose to divid somewhere

Answer 4

wouldn't it be d?

Answer 5

well the power of a power index law says

\[(x^a)^b= x^{a \times b}\]

so what happens on the left hand side

Answer 6

remember you have

\[(2^{-1})^{2x} = \]

Answer 7

you multiply a and b so 2^-2x

Answer 8

thats correct... so on both sides of the equation you have the same base.... 2

so you can just equate the powers

Answer 9

power on the left equals power on the right

Answer 10

do we add the exponents?

Answer 11

no just equate them here is an example... both terms have a base of b

\[b^{3a} = b^m ~~~~then~~~~~3a = m\]

Answer 12

so -2=5

Answer 13

-2x=5

Answer 14

that's correct

Answer 15

Thanks so so much. I though no one would help me. I fanned and medal you. Thanks again