Question

You have two exponential functions. One function has the formula g(x) = 3(2 x ). The other function has the formula h(x) = 2 x+1. Which option below gives formula for k(x) = (g – h)(x)? k(x) = 2x k(x) = 5(2x) k(x) = 5(2x+1) k(x) = 2

Answer 1

`k(x)=2^(x)` ⇒ 1st answer

Explanation:

* Lets explain how to solve the problem

∵
`g(x)=3(2^(x))`

∵
`h(x)=2^(x+1)`

- Lets revise this rule to use it

# If
`a^(n)*a^(m)=a^(n+m)====then==== a^(n+m)=a^(n)*a^(m)`

- We will use this rule in h(x)

∵
`h(x)=2^(x+1)`

- Let a = 2 , n = x , m = 1

∴
`h(x)=2^(x)*2^(1)`

- Now lets find k(x)

∵ k(x) = (g - h)(x)

∵
`g(x)=3(2^(x))`

∵
`h(x)=2^(x)*2^(1)`

∴
`k(x)=3(2^(x))-(2^(x)*2^(1))`

- We have two terms with a common factor
`2^(x)`

∵
`2^(x)` is a common factor

∵
`(3(2^(x)))/(2^(x))=3`

∵
`(2^(x)*2^(1))/(2^(x))=2^(1)=2`

∴
`k(x) = 2^(x)[3 - 2]=2^(x)(1)=2^(x)`

*
`k(x)=2^(x)`