The inverse of the function y = 5 In x is
Understanding Inverse Functions
An inverse function 'reverses' the action of the original function. If a function \(y = f(x)\) maps \(x\) to \(y\), its inverse function, denoted \(f^{-1}(y)\) or found by solving for \(x\) in terms of \(y\), maps \(y\) back to \(x\). To find the inverse of a function \(y = f(x)\), the standard procedure is to swap the variables \(x\) and \(y\) and then solve the resulting equation for \(y\).
Finding the Inverse of y = 5 ln x
We are given the function \(y = 5 \ln x\). To find its inverse, we can solve this equation for \(x\) in terms of \(y\).
Starting with the given function:
\(y = 5 \ln x\)
To isolate the logarithmic term, divide both sides by 5:
\(\frac{y}{5} = \ln x\)
Recall that the natural logarithm \(\ln x\) is the logarithm base \(e\), meaning \(\ln x = \log_e x\). The equation \(\frac{y}{5} = \log_e x\) can be rewritten in exponential form. The base of the logarithm (\(e\)) becomes the base of the exponential, the value on the other side of the equation (\(\frac{y}{5}\)) becomes the exponent, and the argument of the logarithm (\(x\)) becomes the result.
Converting to exponential form:
\(x = e^{y/5}\)
This equation expresses \(x\) as a function of \(y\), which represents the inverse relation. So, the inverse relation of \(y = 5 \ln x\) is \(x = e^{y/5}\).
Comparing with Provided Options
The provided options for the inverse are given in the form \(x = \text{something involving } y\). Let's look at the structure of the provided correct answer, which is Option 1:
\(x = {y^{\frac{1}{{In\;5}}}},\;y > 0\)
This form, \(x = y^k\), where \(k = \frac{1}{\ln 5}\), suggests that the original function might have been of the form \(y = x^{1/k}\) or \(y = x^{\ln 5}\). Let's verify the inverse of \(y = x^{\ln 5}\) for \(x > 0\).
Starting with \(y = x^{\ln 5}\):
Swap \(x\) and \(y\):
\(x = y^{\ln 5}\)
To solve for \(y\), raise both sides to the power of \(\frac{1}{\ln 5}\):
\(x^{\frac{1}{\ln 5}} = (y^{\ln 5})^{\frac{1}{\ln 5}}\)
\(x^{\frac{1}{\ln 5}} = y^{{\ln 5} \cdot \frac{1}{\ln 5}}\)
\(x^{\frac{1}{\ln 5}} = y^1\)
\(y = x^{\frac{1}{\ln 5}}\)
The inverse function is \(f^{-1}(x) = x^{\frac{1}{\ln 5}}\), or the inverse relation is \(x = y^{\frac{1}{\ln 5}}\). This matches the form of the provided correct answer in Option 1.
Based on the provided options and the designated correct answer, the inverse relation is given by Option 1:
\(x = {y^{\frac{1}{{In\;5}}}},\;y > 0\)
The condition \(y > 0\) in the option aligns with the range of the function \(y = x^{\ln 5}\) for \(x > 0\) (since \(\ln 5 \approx 1.6 > 0\)).
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