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?

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Answer 1 · 2018-05-03T11:21:23

The range is #(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)#