Question

If `f_1(x) = 2^(f_2(x))`, where `f_2(x) = 2012^(f_3(x))`, where `f_3(x) = (1/2013)^(f_4(x))`, where `f_4(x) = log_2013 log_x 2012`, then the range of `f_1(x)` is?

Answer 1

The range is `(2,oo)`

Explanation:

We know that, for `x in RR^+,n inRR,and a inRR^+ -{1}`,then

`color(red)((1)log_a x^n=nlog_a x`

`color(blue)((2)B^( (log_B p) )=p...to p inRR^+`

`color(violet)((3)log_a b=1/(log_b a)`

We have,

`f_4(x)=log_2013 (log_x 2012)`

`f_4(x)=log_2013 M,where,M=log_x 2012`

Now ,

`f_3(x)=(1/2013)^(f_4(x))`

`f_3(x)=(2013)^(-f_4(x))...to[ as, a^(-n)=1/(a^n)]`

`f_3(x)=(2013)^(-log_2013 M)`

`f_3(x)=(2013)^(log_2013 M^-1)...to[color(red)(use (1))]`

`f_3(x)=M^-1...tocolor(blue)([use (2),with,B=2013and p=M^-1]`

`f_3(x)=1/M, where, M=log_x 2012`

`f_3(x)=1/(log_x 2012),...to[color(violet)(use (3))]`

`f_3(x)=log_2012 x`

Now,we take

`f_2(x)=(2012)^(f_3(x))`

`f_2(x)=(2012)^(log_2012 x)`

`f_2(x)=x...to [color(blue)(use (2))]`

So,

`f_1(x)=2^(f_2(x))`

`f_1(x)=2^x`

Hence, `f_1` is an Exponential Function.

In general the domain of `color(red)(2^x` is `RR` and range of `color(red)(2^x` is `RR^+=color(red)((0,oo)`

But, for # color(blue)(log_x 2012 > 0, x > 1=>#Range of `color(red)(f_1` is `color(blue)((2,oo)`