exponential functions

[tsh44]

hey i do not know how how to do this...

for the exponential function f (x) =ab^x, suppsoe f (2) = 2 and f (4) = 18

I have to find a & b and f^-1(54)

 

[Guest]

f (x) =ab^x,
f (2) = 2 and f (4) = 18

f(2) means place the value x= 2 into the equation

f(2) = ab^2 and this is also equal to 2

f(4) = ab^4 and this is equal to 18

now you have 2 equations with 2 unknowns to solve....

 

[xc630]

Ok I have (ab)^2=2 and (ab)^4= 18. How do I find a&b if there are two unknowns in each equation?

 

[Guest]

(ab)^2=2 and (ab)^4= 18

rewrite then as this,

(ab) = V2 (where V is the square root)

a= (V 2)/b
not place this back ino the other equation so you only have b as the unknown. Then solve for b.

 

[tsh44]

I got b to be 9 SQRT 2. Is this right?

 

[Guest]

I assume you have written part of the problem in error, but I wanted to focus on the method of solution. Maybe your (ab)^2 is (a (b^2)) or something similar.

If you look at the following logic....
(ab)^2=2 and (ab)^4= 18

then ((ab)^2)^2) = (ab)^4

then use the replacement for (ab)^2 = 2 and (ab)^4 = 18
and you get....
2^2 =18
4=18...and this is rubbish.

 

[xc630]

uum ?