Let f(x) = log_b x, and let g(x) = x^2 - 4x + 4. given that f(g(x)) = g(f(x)) = 0 has exactly one solution and that b > 1, compute b.

Guest Jul 19, 2019

#1**+2 **

f(x) = log_{b} x g(x) = x^2 -4x + 4 f(g) = 0 and g(f) = 0

So

f(g) = log_{b} [ x^2 -4x + 4] = 0

So....in exponential form

b^0 = x^2 -4x +4

1 = x^2 -4x + 4

x^2 - 4x +3 = 0

(x -3) ( x -1) =0

x = 3 or x =1

And

g(f) = [ log_{b} x]^2 - 4[log_{b} x ] + 4 = 0 [remember that log_{ b} 1 = 0 ]....so.....

If x = 1 ⇒ [log _{b} 1 ]^2 - 4[ log _{b} 1] + 4 ⇒ [0]^2 - 4 [0 ] + 4 = 4

So....x =1 doesn't lead to a solution

If x = 3

[ log _{b} 3 ]^2 - 4 [log_{ b} 3 ] + 4 = 0 ....... and by a log property

2log_{b} 3 - 4 log_{ b} 3 + 4 = 0

-2logb 3 + 4 = 0

4 = 2 log_{ b} 3

2 = log _{b} 3

In exponential form

b^2 = 3 and since b > 1

b = √3

CPhill Jul 19, 2019