Question

F(x) = log 2(x + 7) and g(x) = log 2(3x + 5).

(a) Solve f(x) = 4 What point is on the graph of f?
(b) Solve g(x) = 3. What point is on the graph of g?
(c) Solve f(x) = g(x). Do the graphs off and g intersect? If so, where?
(d) Solve (f + g)(x) = 9
(e) Solve (f-9)(x) = 3

Answer 1

(a) x = 9, (9,4)

(b) x = 1, (1,3)

(c) x = 1 and the graph of f(x) and g(x) intersects at point (1,3)

(d)
`x = - (53)/(3)` or x = 9

(e)
`x = - (33)/(23)`

Explanation:

We are given that
`f(x) = \log_(2) {(x + 7)}` ....... (1),and

`g(x) = \log _(2) {(3x + 5)}` ........ (2)

Now,

(a) We have to solve f(x) = 4

⇒
`f(x) = \log_(2) {(x + 7)} = 4`

Converting logarithm to exponent form, we get,

`x + 7 = 2^(4) = 16`

⇒ x = 9 (Answer)

Now, the point on the graph of f(x) will be (9,4) (Answer)

(b) We have to solve g(x) = 3

⇒
`g(x) = \log_(2) {(3x + 5)} = 3`

Converting logarithm to exponent form, we get,

`3x + 5 = 2^(3) = 8`

⇒ x = 1 (Answer)

Now, the point on the graph of g(x) will be (1,3) (Answer)

(c) We have to solve f(x) = g(x)

⇒
`\log_(2) {(x + 7)} = \log _(2) {(3x + 5)}`

Now comparing both sides we can write

x + 7 = 3x + 5

⇒ 2x = 2

⇒ x = 1 (Answer)

Now, at x = 1,
`f(x) = \log_(2) {(1 + 7)} = \log_(2) {2^(3)} = 3`

So, the graph of f(x) and g(x) intersects at point (1,3) (Answer)

(d) We have to solve (f + g)(x) = 9

⇒
`\log_(2) {(x + 7)} + \log _(2) {(3x + 5)} = 9`

⇒
`\log_(2) {(x + 7)(3x + 5)} = 9`

⇒
`(x + 7)(3x + 5) = 2^(9) = 512`

⇒ 3x² + 26x - 477 = 0

⇒ (3x + 53)(3x - 27) = 0

Hence,
`x = - (53)/(3)` or x = 9 (Answer)

(e) We have to solve (f - g)(x) = 3

⇒
`\log_(2) {(x + 7)} - \log _(2) {(3x + 5)} = 3`

⇒
`\log_(2) {(x + 7)/(3x + 5)  = 3`

⇒
`(x + 7)/(3x + 5) = 2^(3) = 8`

⇒ x + 7 = 24x + 40

⇒ 23x = - 33

⇒
`x = - (33)/(23)` (Answer)