If $f_{1} (x) = 2^{f_{2} (x)}$ $f_1(x) = 2^(f_2(x))$ , where $f_{2} (x) = 2012^{f_{3} (x)}$ $f_2(x) = 2012^(f_3(x))$ , where $f_{3} (x) = {(\frac{1}{2013})}^{f_{4} (x)}$ $f_3(x) = (1/2013)^(f_4(x))$ , where $f_{4} (x) = {log}_{2013} {log}_{x} 2012$ $f_4(x) = log_2013 log_x 2012$ , then the range of $f_{1} (x)$ $f_1(x)$ is?

Question

1648 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-03T11:21:23

The range is $(2, \infty)$ $(2,oo)$

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 $color(blue)((2,oo)$