Help please!!!!

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Help please!!!!

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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what do u need help with
yes im thinkin
taking note

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can u add the exponents?
add them
Can you help?
yes im tryin my best
it would be 4^3=2 i think
srry if i am no help
You are but I was seeing a few examples of how to do this, and that doesn't look right to me.
okay what do u think it is
4 = 2^2 so 4^-1 = 2^(-2)
|dw:1443113530353:dw|
as the bases are both 2 we can say -2x = x + 3
so just solve that equation and you have your answer
oh okay, but I have to make a table for these two equations with integer values between 3 and -3
oh so you just plug in values form -3 to 3 so if x = -3 4^(--3) = 4^3 = 64 and 2^(-3+3) = 2^0 = 1
when x = -1 you'll find they both have the value 1/4
- that is the solution of the above equation
did you understand how i got 64 and 1 for when x = -3?
Yes I do sir.
ok then lets try x = -2 4^-x = 4^(-(-2) = 4^2 = 16 2^(x+3) = 2^(-2+3) = 2^1 = 2
note when the power becomes negative you convert to 1 / the postive value for example 4^-3 = 1/4^3 = 1/64
oh okay I got it now.
Thank you very much!
yw . You can do it. You are fearless!
@welshfella can you tell me how to do the rest and if any of mine are incorrect.
@Nnesha , @welshfella can you tell me if I did any of these wrong and how to solve the rest.
yes the top 3 are correct what about value of 2^(x+3) when x = 0?
sorry gotta go
I can't figure out the rest.
if you have \[2^{x + 3}\] when x = 0 you have \[2^{0 + 3} = 2^3 = \] just calculate the answer now
correct... and there are a few errors looking at \[4^{-x}\] when x = -3 its \[4^{-(-3)} = 4^3 = \] calculate a value..
okay so would it be 64 right? And it has to be in a fraction form right?
that's correct... the column 4^(-x) you start with 64 and then divide by 4 |dw:1443134411663:dw| snf 2^{x + 3) you are multiplying the previous answer by 2 hope it makes sense
|dw:1443134586802:dw| that should help
looks perfect

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