Exponential Function f(x) = ab^x w/ f(2) = 2 and f(4) = 18

[ptsoccerboy]

For the exponential function f(x) = ab^x, suppose f(2) = 2 and f(4) = 18

a. Find a and b

f(2) = ab^2
2 = ab^2
a = (2 / b^2)

f(4) = ab^4
18 = ab^4
(18 / a) = b^4
Fourth root(18 / a) = b

I do not know how to find the numerical value of a and b.

b. Find f^-1(54)

f(x) = ab^x
y = ab^x
x = ab^y
logx = ylogab
y = (logx / logab)
f^-1(x) = (logx / logab)

f^-1(54) = (log54/.......)

I do not know how to figure out this problem. Please help me. Thank you!

 

[Deleted member 4993]

Re: Exponential Functions

For the exponential function f(x) = ab^x, suppose f(2) = 2 and f(4) = 18

a. Find a and b

f(2) = ab^2
2 = ab^2
a = (2 / b^2)

f(4) = ab^4
18 = ab^4
(18 / a) = b^4
Fourth root(18 / a) = b

2 = a * b^2...................................(1)

18 = a*b^4...................................(2)

Divide (2) by (1) - and solve for 'b'.

Then use (1) and solve for 'a'

b. Find f^-1(54)

f(x) = ab^x
y = ab^x
x = ab^y
logx = ylogab.........................Incorrect log (x) = log (a) + y * log(b)

 

[galactus]

You have:

\(\displaystyle \L\\ab^{2}=2\)........[1]
\(\displaystyle \L\\ab^{4}=18\)......[2]

Solve [1] for a:

\(\displaystyle \L\\a=\frac{2}{b^{2}}\)

Sub into [2]:

\(\displaystyle \L\\\frac{2}{b^{2}}\cdot{b^{4}}=18\)

\(\displaystyle \L\\2b^{2}=18\)

\(\displaystyle \L\\b=3\)

Now, resub to find a.

 

[ptsoccerboy]

when u divide (2) by (1)
you're supposed to get 9 = b^2 correct?

 

[ptsoccerboy]

oh ok thank you very much!

 

[Deleted member 4993]

when u divide (2) by (1)
you're supposed to get 9 = b^2 correct?

so

b = ±3

If b < 0 then the function only exists for all integer number (in real domain).