# 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?

May 3, 2018

The range is $\left(2 , \infty\right)$

#### Explanation:

We know that, for $x \in {\mathbb{R}}^{+} , n \in \mathbb{R} , \mathmr{and} a \in {\mathbb{R}}^{+} - \left\{1\right\}$,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,

${f}_{4} \left(x\right) = {\log}_{2013} \left({\log}_{x} 2012\right)$

${f}_{4} \left(x\right) = {\log}_{2013} M , w h e r e , M = {\log}_{x} 2012$

Now ,

${f}_{3} \left(x\right) = {\left(\frac{1}{2013}\right)}^{{f}_{4} \left(x\right)}$

${f}_{3} \left(x\right) = {\left(2013\right)}^{- {f}_{4} \left(x\right)} \ldots \to \left[a s , {a}^{- n} = \frac{1}{{a}^{n}}\right]$

${f}_{3} \left(x\right) = {\left(2013\right)}^{- {\log}_{2013} M}$

${f}_{3} \left(x\right) = {\left(2013\right)}^{{\log}_{2013} {M}^{-} 1} \ldots \to \left[\textcolor{red}{u s e \left(1\right)}\right]$

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

${f}_{3} \left(x\right) = \frac{1}{M} , w h e r e , M = {\log}_{x} 2012$

${f}_{3} \left(x\right) = \frac{1}{{\log}_{x} 2012} , \ldots \to \left[\textcolor{v i o \le t}{u s e \left(3\right)}\right]$

${f}_{3} \left(x\right) = {\log}_{2012} x$

Now,we take

${f}_{2} \left(x\right) = {\left(2012\right)}^{{f}_{3} \left(x\right)}$

${f}_{2} \left(x\right) = {\left(2012\right)}^{{\log}_{2012} x}$

${f}_{2} \left(x\right) = x \ldots \to \left[\textcolor{b l u e}{u s e \left(2\right)}\right]$

So,

${f}_{1} \left(x\right) = {2}^{{f}_{2} \left(x\right)}$

${f}_{1} \left(x\right) = {2}^{x}$

Hence, ${f}_{1}$ is an Exponential Function.

In general the domain of color(red)(2^x is $\mathbb{R}$ 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)