Question

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 1

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.