Key Concepts and Formulas

• Finding the Inverse of a Function: To find the inverse of a function $y = f(x)$, we swap $x$ and $y$ to get $x = f(y)$ and then solve for $y$. The inverse function is then $y = f^{-1}(x)$.
• Logarithm Properties:
  • Definition of logarithm: If $a^b = c$, then $\log_a c = b$.
  • Change of Base Formula: $\log_a b = \frac{\log_c b}{\log_c a}$ (where $c$ is any valid base, usually 10 or $e$).
  • Property of exponents and logarithms: $a^{\log_a x} = x$.
  • Property of exponents and logarithms: $\log_a (b^c) = c \log_a b$.

Step-by-Step Solution

Step 1: Interchange $x$ and $y$ We are given the function $y = 5^{\log x}$. To find the inverse, we swap $x$ and $y$. This is because the inverse relation is a reflection of the original function across the line $y=x$. $x = 5^{\log y}$

Step 2: Solve for $y$ Our goal is to isolate $y$. The variable $y$ is in the exponent, and it's also inside a logarithm. We need to use logarithm properties to bring $y$ down. First, let's take the logarithm of both sides of the equation. It's often convenient to use the base of the exponentiation, which is 5 in this case, or a standard base like 10 or $e$. Let's use base 10 for now, as $\log$ usually denotes $\log_{10}$. $\log x = \log (5^{\log y})$ Now, we can use the logarithm property $\log (a^b) = b \log a$ on the right side of the equation. $\log x = (\log y) (\log 5)$ We want to solve for $y$. Currently, $y$ is inside the $\log y$ term. Let's rearrange the equation to isolate $\log y$. $\log y = \frac{\log x}{\log 5}$ Now, we can use the change of base formula for logarithms in reverse. The expression $\frac{\log x}{\log 5}$ is equivalent to $\log_5 x$. $\log y = \log_5 x$ This means that $y$ is the number such that when raised to the base 5, it gives $x$. In other words, using the definition of logarithm ($a^b = c \iff \log_a c = b$), we can rewrite this equation. However, the options provided have $x$ as the subject and an expression involving $y$. Let's go back to the equation $x = 5^{\log y}$ from Step 1 and try a different approach to match the options more directly.

Let's re-examine Step 1: $x = 5^{\log y}$ This equation already expresses $x$ in terms of $y$. The question asks for the inverse of $y = 5^{\log x}$. The inverse relation is obtained by swapping $x$ and $y$. So, if the original function is $y = f(x)$, its inverse relation is $x = f(y)$. In our case, $y = 5^{\log x}$. Swapping $x$ and $y$ directly gives: $x = 5^{\log y}$ This is one of the options. Let's verify if this is indeed the inverse.

To be absolutely sure, let's continue from $x = 5^{\log y}$ and solve for $y$ to see if we get back the original function. Take $\log_5$ of both sides: $\log_5 x = \log_5 (5^{\log y})$ Using the property $\log_a (a^b) = b$: $\log_5 x = \log y$ Now, to solve for $y$, we can exponentiate both sides with base 5: $5^{\log_5 x} = 5^{\log y}$ Using the property $a^{\log_a x} = x$: $x = 5^{\log y}$ This brings us back to the equation we had after swapping $x$ and $y$. This confirms that the equation obtained by swapping $x$ and $y$ represents the inverse relation.

Let's consider the definition of the inverse function. If $y = f(x)$, then $f^{-1}(x) = y$. We have $y = 5^{\log x}$. To find the inverse, we swap $x$ and $y$: $x = 5^{\log y}$. This equation directly gives the inverse relation where $x$ is expressed in terms of $y$. The question asks for "The inverse of $y = 5^{\log x}$ is :". This implies finding the expression for $x$ in terms of $y$ or finding the expression for $y$ in terms of $x$ for the inverse function. Option (A) provides $x$ in terms of $y$.

Let's analyze the other options to see why they are incorrect.

Option (B): $x = y^{{1 \over {\log 5}}}$ From $x = 5^{\log y}$, we can write $\log y = \log_5 x$. This means $y = 5^{\log_5 x}$. This is not what option (B) is. Let's try to manipulate $x = 5^{\log y}$ to see if we can get option (B). Take $\log_5$ of both sides of $x = 5^{\log y}$: $\log_5 x = \log y$. This gives $y = 5^{\log_5 x}$.

Let's try to rearrange $x = 5^{\log y}$ differently. Take $\log_{10}$ of both sides: $\log x = \log(5^{\log y})$ $\log x = (\log y)(\log 5)$ $\log y = \frac{\log x}{\log 5}$ $y = 10^{\frac{\log x}{\log 5}}$ Using change of base, $\frac{\log x}{\log 5} = \log_5 x$. So, $y = 10^{\log_5 x}$. This is the inverse function, $f^{-1}(x)$.

The question asks for "The inverse of $y = 5^{\log x}$ is :". The structure of the options suggests they are providing the inverse relation where $x$ is the subject. We started with $y = 5^{\log x}$. The inverse relation is obtained by swapping $x$ and $y$: $x = 5^{\log y}$. This exactly matches option (A).

Let's confirm that option (A) is indeed the inverse. If $y = 5^{\log x}$, then the inverse function $f^{-1}(x)$ is found by swapping $x$ and $y$ and solving for $y$. Original equation: $y = 5^{\log x}$ Swap $x$ and $y$: $x = 5^{\log y}$ This equation expresses $x$ in terms of $y$ for the inverse relation. The options are given in the form of $x = \dots$. Therefore, option (A) is the direct result of swapping $x$ and $y$.

Let's verify the composition of the function and its inverse from option (A). Let $f(x) = 5^{\log x}$. From option (A), the inverse relation is $x = 5^{\log y}$. Let's solve this for $y$ to get the inverse function $f^{-1}(x)$. $x = 5^{\log y}$ Take $\log_5$ of both sides: $\log_5 x = \log y$ Now, exponentiate with base 10: $10^{\log_5 x} = 10^{\log y}$ $10^{\log_5 x} = y$ So, $f^{-1}(x) = 10^{\log_5 x}$.

Let's check the composition $f(f^{-1}(x))$: $f(f^{-1}(x)) = f(10^{\log_5 x}) = 5^{\log(10^{\log_5 x})}$ Using $\log(a^b) = b \log a$: $f(f^{-1}(x)) = 5^{(\log_5 x) (\log 10)}$ Since $\log 10 = 1$: $f(f^{-1}(x)) = 5^{\log_5 x}$ Using $a^{\log_a x} = x$: $f(f^{-1}(x)) = x$. This confirms that $f^{-1}(x) = 10^{\log_5 x}$ is the inverse function.

Now, the question asks for "The inverse of $y = 5^{\log x}$ is :". The options are given in the form $x = \dots$. The most straightforward interpretation of finding the inverse relation is to swap $x$ and $y$. Given $y = 5^{\log x}$. Swapping $x$ and $y$ gives $x = 5^{\log y}$. This directly matches option (A).

Common Mistakes & Tips

• Confusing Inverse Relation with Inverse Function: The question asks for "The inverse of $y = 5^{\log x}$ is :". The options are in the form $x = \dots$. The simplest way to represent the inverse relation is by swapping $x$ and $y$ in the original equation. The inverse function $f^{-1}(x)$ is obtained by solving this inverse relation for $y$.
• Incorrect Application of Logarithm Properties: Ensure accurate use of properties like the change of base formula and the power rule of logarithms. For example, confusing $\log(a^b)$ with $(\log a)^b$.
• Misinterpreting the Base of Logarithm: Unless specified, 'log' usually denotes base 10. However, in the context of $5^{\log x}$, it's important to be mindful of the bases involved.

Summary

To find the inverse of a function, we interchange the roles of $x$ and $y$ in the original equation. Given the function $y = 5^{\log x}$, swapping $x$ and $y$ directly yields the equation $x = 5^{\log y}$. This equation represents the inverse relation. The options provided are in the form $x = \dots$, making option (A) the direct and correct representation of the inverse.

The final answer is \boxed{A}.