Logarithmic Inversion

Algebra Level 2

If f : ( 0 , ∞ ) → R f : (0,\infty) \to \mathbb R be defined as f ( x ) = log ⁡ 2 x f(x) = \log_{2} x then find f − 1 ( 2 ) f^{-1} (2) .


The answer is 4.

Ram Mohith by Ram Mohith , 18, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Chew-Seong Cheong
Feb 13, 2019

Given that

f ( x ) = log ⁡ 2 x Let x = f − 1 ( u ) ⟹ f ( x ) = f ( f − 1 ( u ) ) = u u = log ⁡ 2 ( f − 1 ( u ) ) Antilogarithm of base 2 on both sides 2 u = f − 1 ( u ) Replace u with x and rearrange f − 1 ( x ) = 2 x ⟹ f − 1 ( 2 ) = 2 2 = 4 \begin{aligned} f(x) & = \log_2 x & \small \color{#3D99F6} \text{Let }x = f^{-1}(u) \implies f(x) = f\left(f^{-1} (u)\right) = u \\ u & = \log_2 \left(f^{-1}(u)\right) & \small \color{#3D99F6} \text{Antilogarithm of base 2 on both sides} \\ 2^u & = f^{-1} (u) & \small \color{#3D99F6} \text{Replace }u \text{ with }x \text{ and rearrange} \\ f^{-1} (x) & = 2^x \\ \implies f^{-1}(2) & = 2^2 = \boxed 4 \end{aligned}

4 Helpful 1 Interesting 1 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...