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.
+128407 f(x) = logb x g(x) = x^2 -4x + 4 f(g) = 0 and g(f) = 0
So
f(g) = logb [ 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) = [ logb x]^2 - 4[logb 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
2logb 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
