Question

The equation
`f(x)=7a^(2+x)-b` has an x-intercept equivalent to

A)
`x=(logb-log7)/(loga)-2`

B)
`x=7a^(2)-b`

C)
`x=(y+b)/(7a^(2))`

D)
`x=0`

I have the answer key so I know the answer. I just don't know how they got to the answer so please show your work.

Answer 1

`\textsf{A)} \quad x=(\log b - \log 7)/( \log a)-2`

Explanation:

Given function:

`f(x)=7a^(2+x)-b`

The x-intercepts occur when the function equals zero.

`\implies 7a^(2+x)-b=0`

Add b to both sides:

`\implies 7a^(2+x)=b`

Divide both sides by 7:

`\implies a^(2+x)=(b)/(7)`

Take logs of both sides of the equation:

`\implies \log \left(a^(2+x)\right)= \log \left((b)/(7)\right)`

`\textsf{Apply the Power log law}: \quad \log_ax^n=n\log_ax`

`\implies (2+x)\log a= \log \left((b)/(7)\right)`

`\textsf{Apply the Quotient log law}: \quad \log_a(x)/(y)=\log_ax - \log_ay`

`\implies (2+x)\log a= \log b-\log 7`

Divide both sides by log a:

`\implies 2+x=(\log b - \log 7)/( \log a)`

Subtract 2 from both sides:

`\implies x=(\log b - \log 7)/( \log a)-2`