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:(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) [tex]x = - \frac{53}{3}[/tex] or x = 9(e) [tex]x = - \frac{33}{23}[/tex]Step-by-step explanation:We are given that [tex]f(x) = \log_{2} {(x + 7)}[/tex] ....... (1),and  [tex]g(x) = \log _{2} {(3x + 5)}[/tex] ........ (2) Now,  (a) We have to solve f(x) = 4 ⇒ [tex]f(x) = \log_{2} {(x + 7)} = 4[/tex] Converting logarithm to exponent form, we get, [tex]x + 7 = 2^{4} = 16[/tex] ⇒ 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 ⇒ [tex]g(x) = \log_{2} {(3x + 5)} = 3[/tex] Converting logarithm to exponent form, we get, [tex]3x + 5 = 2^{3} = 8[/tex] ⇒ 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) ⇒ [tex]\log_{2} {(x + 7)} = \log _{2} {(3x + 5)}[/tex] Now comparing both sides we can write x + 7 = 3x + 5  ⇒ 2x = 2  ⇒ x = 1 (Answer) Now, at x = 1, [tex]f(x) = \log_{2} {(1 + 7)} = \log_{2} {2^{3}} = 3[/tex]  So, the graph of f(x) and g(x) intersects at point (1,3) (Answer) (d) We have to solve (f + g)(x) = 9 ⇒ [tex]\log_{2} {(x + 7)} + \log _{2} {(3x + 5)} = 9[/tex] ⇒  [tex]\log_{2} {(x + 7)(3x + 5)} = 9[/tex] ⇒ [tex](x + 7)(3x + 5) = 2^{9} = 512[/tex] ⇒ 3x² + 26x - 477 = 0 ⇒ (3x + 53)(3x - 27) = 0 Hence, [tex]x = - \frac{53}{3}[/tex] or x = 9 (Answer) (e) We have to solve (f - g)(x) = 3 ⇒ [tex]\log_{2} {(x + 7)} - \log _{2} {(3x + 5)} = 3[/tex] ⇒ [tex]\log_{2} {\frac{x + 7}{3x + 5}  = 3[/tex] ⇒ [tex]\frac{x + 7}{3x + 5} = 2^{3} = 8[/tex] ⇒ x + 7 = 24x + 40 ⇒ 23x = - 33 ⇒ [tex]x = - \frac{33}{23}[/tex] (Answer)