
If \[f\left( x \right) = 5{\log _5}x\], then find the value of \[{f^{ - 1}}\left( {\alpha - \beta } \right)\] where \[\alpha ,\beta \in R\].
A. \[{f^{ - 1}}\left( \alpha \right) - {f^{ - 1}}\left( \beta \right)\]
B. \[\dfrac{{{f^{ - 1}}\left( \alpha \right)}}{{{f^{ - 1}}\left( \beta \right)}}\]
C. \[\dfrac{1}{{f\left( {\alpha - \beta } \right)}}\]
D. \[\dfrac{1}{{f\left( \alpha \right) - f\left( \beta \right)}}\]
Answer
162.9k+ views
Hint: Here, a logarithmic function is given. First, consider the given function as \[y = 5{\log _5}x\]. Then, solve the equation and find the value of \[x\] in terms of \[y\]. After that, calculate the value of \[{f^{ - 1}}\left( x \right)\]. In the end, substitute \[x = \alpha - \beta \] in the equation of \[{f^{ - 1}}\left( x \right)\] and solve it to get the required answer.
Formula Used: \[{e^{{{\log }_e}x}} = x\]
\[\dfrac{{a - b}}{c} = \dfrac{a}{c} - \dfrac{b}{c}\]
Complete step by step solution: The given function is \[f\left( x \right) = 5{\log _5}x\].
Let’s simplify the given function.
Consider,
\[y = 5{\log _5}x\]
\[ \Rightarrow \dfrac{y}{5} = {\log _5}x\]
Now take both sides as the exponent of the number 5.
\[ \Rightarrow {5^{\dfrac{y}{5}}} = {5^{{{\log }_5}x}}\]
Apply the exponent rule of logarithm \[{e^{{{\log }_e}x}} = x\] on the right-hand side.
\[ \Rightarrow {5^{\dfrac{y}{5}}} = x\]
So, we get
\[{f^{ - 1}}\left( x \right) = {5^{\dfrac{x}{5}}}\] \[.....\left( 1 \right)\]
Now substitute \[x = \alpha - \beta \] in the above equation.
\[{f^{ - 1}}\left( {\alpha - \beta } \right) = {5^{\dfrac{{\alpha - \beta }}{5}}}\]
Solve the right-hand side.
\[{f^{ - 1}}\left( {\alpha - \beta } \right) = {5^{\left( {\dfrac{\alpha }{5} - \dfrac{\beta }{5}} \right)}}\]
Apply the exponent rule.
\[{f^{ - 1}}\left( {\alpha - \beta } \right) = {5^{\left( {\dfrac{\alpha }{5}} \right)}} \times {5^{\left( {\dfrac{{ - \beta }}{5}} \right)}}\]
\[ \Rightarrow {f^{ - 1}}\left( {\alpha - \beta } \right) = \dfrac{{{5^{\left( {\dfrac{\alpha }{5}} \right)}}}}{{{5^{\left( {\dfrac{\beta }{5}} \right)}}}}\]
Now compare the right-hand side with the equation \[\left( 1 \right)\].
\[ \Rightarrow {f^{ - 1}}\left( {\alpha - \beta } \right) = \dfrac{{{f^{ - 1}}\left( \alpha \right)}}{{{f^{ - 1}}\left( \beta \right)}}\]
Option ‘B’ is correct
Note: Students often do mistake while solving the functions with logarithm. To simplify the function, take the function as the exponent of a number. Where the number is the base of the given logarithm.
Formula Used: \[{e^{{{\log }_e}x}} = x\]
\[\dfrac{{a - b}}{c} = \dfrac{a}{c} - \dfrac{b}{c}\]
Complete step by step solution: The given function is \[f\left( x \right) = 5{\log _5}x\].
Let’s simplify the given function.
Consider,
\[y = 5{\log _5}x\]
\[ \Rightarrow \dfrac{y}{5} = {\log _5}x\]
Now take both sides as the exponent of the number 5.
\[ \Rightarrow {5^{\dfrac{y}{5}}} = {5^{{{\log }_5}x}}\]
Apply the exponent rule of logarithm \[{e^{{{\log }_e}x}} = x\] on the right-hand side.
\[ \Rightarrow {5^{\dfrac{y}{5}}} = x\]
So, we get
\[{f^{ - 1}}\left( x \right) = {5^{\dfrac{x}{5}}}\] \[.....\left( 1 \right)\]
Now substitute \[x = \alpha - \beta \] in the above equation.
\[{f^{ - 1}}\left( {\alpha - \beta } \right) = {5^{\dfrac{{\alpha - \beta }}{5}}}\]
Solve the right-hand side.
\[{f^{ - 1}}\left( {\alpha - \beta } \right) = {5^{\left( {\dfrac{\alpha }{5} - \dfrac{\beta }{5}} \right)}}\]
Apply the exponent rule.
\[{f^{ - 1}}\left( {\alpha - \beta } \right) = {5^{\left( {\dfrac{\alpha }{5}} \right)}} \times {5^{\left( {\dfrac{{ - \beta }}{5}} \right)}}\]
\[ \Rightarrow {f^{ - 1}}\left( {\alpha - \beta } \right) = \dfrac{{{5^{\left( {\dfrac{\alpha }{5}} \right)}}}}{{{5^{\left( {\dfrac{\beta }{5}} \right)}}}}\]
Now compare the right-hand side with the equation \[\left( 1 \right)\].
\[ \Rightarrow {f^{ - 1}}\left( {\alpha - \beta } \right) = \dfrac{{{f^{ - 1}}\left( \alpha \right)}}{{{f^{ - 1}}\left( \beta \right)}}\]
Option ‘B’ is correct
Note: Students often do mistake while solving the functions with logarithm. To simplify the function, take the function as the exponent of a number. Where the number is the base of the given logarithm.
Recently Updated Pages
JEE Advanced 2021 Physics Question Paper 2 with Solutions

Solutions Class 12 Notes JEE Advanced Chemistry [PDF]

JEE Advanced 2022 Chemistry Question Paper 2 with Solutions

JEE Advanced 2025 Revision Notes for Chemistry Energetics - Free PDF Download

JEE Advanced Marks vs Rank 2025 - Predict IIT Rank Based on Score

JEE Advanced 2022 Maths Question Paper 2 with Solutions

Trending doubts
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

IIT CSE Cutoff: Category‐Wise Opening and Closing Ranks

Top IIT Colleges in India 2025

IIT Fees Structure 2025

IIT Roorkee Average Package 2025: Latest Placement Trends Updates

Other Pages
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations
