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)
