Question

At what point on the graph of \(y=3^{x}+1\) is the tangent line parallel to the line \(y=5 x-1 ?\)

Short answer

The point on the graph of \(y=3^{x}+1\) where the tangent line is parallel to the line \(y=5x-1\) is \(\left(\frac{\ln 5}{\ln 3}, 6\right)\).

Step-by-step solution

Calcuate the Derivative

Firstly, obtain the derivative of the function \(y=3^{x}+1\) which will be used to find points on the function's graph where the tangent line's slope equals 5. The derivative of \(y=3^{x}+1\) is given by \(y'=3^{x} \cdot \ln 3\).

Set the derivative equal to the slope of the line

As the next step, set \(y'=3^{x} \cdot \ln 3\) equal to the slope of the given line, \(y=5x-1\), which is 5. Solve the resulting equation \(3^{x} \cdot \ln 3 = 5\), for \(x\), which gives \(x= \frac{\ln 5}{\ln 3}\).

Find the y-coordinate of the point

Substitute the \(x\)-value from step 2 into the original function \(y=3^{x}+1\) to find the \(y\)-coordinate of the point. This yields \[y = 3^{\frac{\ln 5}{\ln 3}}+1 = 5+1 = 6\]. So, the point is \(\left(\frac{\ln 5}{\ln 3}, 6\right)\).