exponential function passing through (3, 4) and (7, 8)

[FMMurphy]

My problem is to find the exponential function, y = ab^x, whose graph passes through the points (3, 4) and (7, 8). I've tried to work it out. This is what I've done:

. . .4 = ab^3
. . .8 = ab^7

. . .a = 4/b^3

This gives me:

. . .8 = 4/b^3(b^7)

Then would it be:

. . .8 = 4b^4

Then 2= b^4. But what's next?

 

[arthur ohlsten]

my approach is different
y=ab^x
ln y= x ln b +ln a

points [3,4] and [7,8]
ln4=3lnb+lna
ln8=7lnb+lna

subtract eq1 from eq2
ln8 - ln4 =[7-3]lnb
ln[2*4]-ln4=4 ln b
ln2+ln4-ln4=4 ln b
4ln b= ln2
lnb=1/4 ln 2
lnb=ln 2^
b=2^1/4

ln4 = 3 ln 2^1/4 + lna from eq1
ln[2^2] =3/4 ln2 +ln a
2 ln2 - 3/4 ln 2=ln a
5/4 ln 2 =ln a
ln 2^5/4 = ln a
a=2^5/4

Arthur

 

[FMMurphy]

Since I need to substitue would the exponential function that is y=ab^x be as follows: y=2^5/4(2^1/4)^x. Is there anyway to simplify that?

 

[FMMurphy]

is the right answer y=2^5/4(2^1/4)^x? I also got an answer from 8=4b^4 and then 2=b^4. From that I got b=1.189 and substituting in I got a= 2.378. If that's right then I would have y=2.37(1.19)^x

 

[tkhunny]

With two points and two parameters, you should be able to reproduce both points exactly. Substitute and check. If you rounded to two decimal places, you won't quite get back where you started.

 

[soroban]

Hello, FMMurphy!

Find the exponential function, \(\displaystyle y \:= \:ab^x\),
whose graph passes through the points \(\displaystyle (3, 4)\) and \(\displaystyle (7, 8)\).

With exponential functions, division is often the neatest way . . .

We have: \(\displaystyle \:\begin{array}{cc}(3,4):\;ab^3\,=\,4 &\;[1]\\ (7,8):\;ab^7 \,=\,8 & \;[2]\end{array}\)

Divide [2] by [1]: \(\displaystyle \L\:\frac{ab^7}{ab^3}\:=\:\frac{8}{4}\)\(\displaystyle \;\;\Rightarrow\;\;b^4\,=\,2\;\;\Rightarrow\;\;b \,=\,\sqrt[4]{2}\)

Substitute into [1]: \(\displaystyle \:a\left(2^{\frac{1}{4}}\right)^3\:=\:4\;\;\Rightarrow\;\;a\,\cdot\,2^{\frac{3}{4}} \:=\:2^2\)

. . Hence: \(\displaystyle \:a\,=\,2^{\frac{5}{4}}\)


The function becomes: \(\displaystyle \:y \:=\:2^{\frac{5}{4}}\,\cdot\left(2^{\frac{1}{4}}\right)^x \;=\;2^{\frac{5}{4}}\,\cdot2^{\frac{x}{4}}\)

\(\displaystyle \text{Therefore: }\L\:y \;=\;2^{\frac{x+5}{4}}\)