Help please!!!!

- anonymous

Help please!!!!

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- schrodinger

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- james1769

what do u need help with

- james1769

yes im thinkin

- james1769

taking note

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## More answers

- james1769

can u add the exponents?

- james1769

add them

- anonymous

Can you help?

- james1769

yes im tryin my best

- james1769

@XxPrisillaXx

- james1769

it would be 4^3=2 i think

- james1769

srry if i am no help

- anonymous

You are but I was seeing a few examples of how to do this, and that doesn't look right to me.

- james1769

okay what do u think it is

- anonymous

@welshfella

- welshfella

4 = 2^2
so
4^-1 = 2^(-2)

- welshfella

|dw:1443113530353:dw|

- welshfella

as the bases are both 2 we can say
-2x = x + 3

- welshfella

so just solve that equation and you have your answer

- anonymous

oh okay, but I have to make a table for these two equations with integer values between 3 and -3

- welshfella

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

- welshfella

when x = -1 you'll find they both have the value 1/4

- welshfella

- that is the solution of the above equation

- welshfella

did you understand how i got 64 and 1 for when x = -3?

- anonymous

Yes I do sir.

- welshfella

ok then lets try x = -2
4^-x = 4^(-(-2) = 4^2 = 16
2^(x+3) = 2^(-2+3) = 2^1 = 2

- welshfella

note when the power becomes negative you convert to 1 / the postive value
for example
4^-3 = 1/4^3 = 1/64

- anonymous

oh okay I got it now.

- anonymous

Thank you very much!

- welshfella

yw . You can do it. You are fearless!

- anonymous

@welshfella can you tell me how to do the rest and if any of mine are incorrect.

- anonymous

@Nnesha , @welshfella can you tell me if I did any of these wrong and how to solve the rest.

- anonymous

@Nnesha

- welshfella

yes the top 3 are correct
what about value of 2^(x+3) when x = 0?

- welshfella

sorry gotta go

- anonymous

@campbell_st

- anonymous

@mathstudent55

- anonymous

I can't figure out the rest.

- campbell_st

if you have
\[2^{x + 3}\]
when x = 0 you have
\[2^{0 + 3} = 2^3 = \]
just calculate the answer now

- campbell_st

correct... and there are a few errors
looking at
\[4^{-x}\]
when x = -3 its
\[4^{-(-3)} = 4^3 = \]
calculate a value..

- anonymous

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

- campbell_st

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

- campbell_st

|dw:1443134586802:dw|
that should help

- campbell_st

looks perfect

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