Question

Help with exponential functions

Answer 1

Bah I lost it all...damn internet

Answer 2

so we will have ab^5=96

Answer 3

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

Answer 4

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?

Answer 5

\(\dfrac{ab^7}{ab^5}=\dfrac{384}{96}\implies b^2 = 4\implies b = 2\)

not very hard at all

Answer 6

\(\dfrac{ab^7}{ab^5}=\dfrac{384}{96}\implies \dfrac{\cancel{a}b^7}{\cancel{a}b^5}=\dfrac{384}{96}\implies \dfrac{b^7}{b^5}=\dfrac{384}{96}\implies b^{7-5}=\dfrac{384}{96}\\\implies b^2 = \dfrac{384}{96}\implies b^2 = 4\implies b = 2\)

Answer 7

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

Answer 8

\(\dfrac{ab^7}{ab^5}=\dfrac{384}{96}\implies \dfrac{\cancel{a}b^7}{\cancel{a}b^5}=\dfrac{384}{96}\implies \dfrac{b^7}{b^5}=\dfrac{384}{96}\implies b^{7-5}=\dfrac{384}{96}\\\implies b^2 = \dfrac{384}{96}\implies b^2 = 4\implies b = 2\)

Answer 9

maybe now if b =2 we can grab a ?

Answer 10

We don't consider \(b=-2\) because we are told \(b\) is positive.

Now we need \(a\), and so we use \(b\)

\(f(5) = 96 \implies a(2)^5 = 96 \implies a32=96 \implies a=3\)

Answer 11

So \(f(x) = 3(2)^x\)

Answer 12

so a2^5=96
2*2*2*2*2 =32
a= 96/32=3
a=3

Answer 13

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

Answer 14

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

Answer 15

correct @DarkBlueChocobo