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DarkBlueChocobo

  • one year ago

Help with exponential functions

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  1. DarkBlueChocobo
    • one year ago
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  2. zzr0ck3r
    • one year ago
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    Bah I lost it all...damn internet

  3. DarkBlueChocobo
    • one year ago
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    so we will have ab^5=96

  4. zzr0ck3r
    • one year ago
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    \(\dfrac{ab^7}{ab^5}=\dfrac{384}{96}\) Solve for \(b\) and then use that to find \(a\).

  5. DarkBlueChocobo
    • one year ago
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    I don't know why that sounds more difficult than it feels like it will be. How would you find B? seperate it or like fill 0 in for a?

  6. zzr0ck3r
    • one year ago
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    \(\dfrac{ab^7}{ab^5}=\dfrac{384}{96}\implies b^2 = 4\implies b = 2\) not very hard at all

  7. zzr0ck3r
    • one year ago
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    \(\dfrac{ab^7}{ab^5}=\dfrac{384}{96}\implies \dfrac{\cancel{a}b^7}{\cancel{a}b^5}=\dfrac{384}{96}\implies \dfrac{b^7}{b^5}=\dfrac{384}{96}\implies b^{7-5}=\dfrac{384}{96}\\\implies b^2 = \dfrac{384}{96}\implies b^2 = 4\implies b = 2\)

  8. DarkBlueChocobo
    • one year ago
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    Oh so a's cancel out and you get left with b's then b^2 = 4 meaning b=2 then cause 384/96 is 4

  9. zzr0ck3r
    • one year ago
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    \(\dfrac{ab^7}{ab^5}=\dfrac{384}{96}\implies \dfrac{\cancel{a}b^7}{\cancel{a}b^5}=\dfrac{384}{96}\implies \dfrac{b^7}{b^5}=\dfrac{384}{96}\implies b^{7-5}=\dfrac{384}{96}\\\implies b^2 = \dfrac{384}{96}\implies b^2 = 4\implies b = 2\)

  10. UsukiDoll
    • one year ago
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    maybe now if b =2 we can grab a ?

  11. zzr0ck3r
    • one year ago
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    We don't consider \(b=-2\) because we are told \(b\) is positive. Now we need \(a\), and so we use \(b\) \(f(5) = 96 \implies a(2)^5 = 96 \implies a32=96 \implies a=3\)

  12. zzr0ck3r
    • one year ago
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    So \(f(x) = 3(2)^x\)

  13. DarkBlueChocobo
    • one year ago
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    so a2^5=96 2*2*2*2*2 =32 a= 96/32=3 a=3

  14. zzr0ck3r
    • one year ago
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    It's just like when you had two linear equations in two variables and you used the addition method. Except instead of adding the equations, we divide them. The \(a, b>0\) gives that we will never divide by zero

  15. UsukiDoll
    • one year ago
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    let me try ... just this once f(x) = ab^x so when f(5) = 96 so x = 5 and b = 2 f(5) =a2^5 96=32a 3 = a

  16. zzr0ck3r
    • one year ago
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    correct @DarkBlueChocobo

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