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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.

 Jul 19, 2019
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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

 

cool cool cool

 Jul 19, 2019

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