anonymous
  • anonymous
Help please!!!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
james1769
  • james1769
what do u need help with
james1769
  • james1769
yes im thinkin
james1769
  • james1769
taking note

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james1769
  • james1769
can u add the exponents?
james1769
  • james1769
add them
anonymous
  • anonymous
Can you help?
james1769
  • james1769
yes im tryin my best
james1769
  • james1769
@XxPrisillaXx
james1769
  • james1769
it would be 4^3=2 i think
james1769
  • james1769
srry if i am no help
anonymous
  • anonymous
You are but I was seeing a few examples of how to do this, and that doesn't look right to me.
james1769
  • james1769
okay what do u think it is
anonymous
  • anonymous
@welshfella
welshfella
  • welshfella
4 = 2^2 so 4^-1 = 2^(-2)
welshfella
  • welshfella
|dw:1443113530353:dw|
welshfella
  • welshfella
as the bases are both 2 we can say -2x = x + 3
welshfella
  • welshfella
so just solve that equation and you have your answer
anonymous
  • anonymous
oh okay, but I have to make a table for these two equations with integer values between 3 and -3
welshfella
  • 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
  • welshfella
when x = -1 you'll find they both have the value 1/4
welshfella
  • welshfella
- that is the solution of the above equation
welshfella
  • welshfella
did you understand how i got 64 and 1 for when x = -3?
anonymous
  • anonymous
Yes I do sir.
welshfella
  • welshfella
ok then lets try x = -2 4^-x = 4^(-(-2) = 4^2 = 16 2^(x+3) = 2^(-2+3) = 2^1 = 2
welshfella
  • welshfella
note when the power becomes negative you convert to 1 / the postive value for example 4^-3 = 1/4^3 = 1/64
anonymous
  • anonymous
oh okay I got it now.
anonymous
  • anonymous
Thank you very much!
welshfella
  • welshfella
yw . You can do it. You are fearless!
anonymous
  • anonymous
@welshfella can you tell me how to do the rest and if any of mine are incorrect.
anonymous
  • anonymous
@Nnesha , @welshfella can you tell me if I did any of these wrong and how to solve the rest.
anonymous
  • anonymous
@Nnesha
welshfella
  • welshfella
yes the top 3 are correct what about value of 2^(x+3) when x = 0?
welshfella
  • welshfella
sorry gotta go
anonymous
  • anonymous
@campbell_st
anonymous
  • anonymous
@mathstudent55
anonymous
  • anonymous
I can't figure out the rest.
campbell_st
  • 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
  • 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
  • anonymous
okay so would it be 64 right? And it has to be in a fraction form right?
campbell_st
  • 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
  • campbell_st
|dw:1443134586802:dw| that should help
campbell_st
  • campbell_st
looks perfect

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