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what do u need help with

yes im thinkin

taking note

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

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

okay so would it be 64 right? And it has to be in a fraction form right?

|dw:1443134586802:dw|
that should help

looks perfect