Question

find a number a such that the graphs of the two functions defined by the rules y = a^x and y = log a x have a common tangent line.

Answer 1

One value of a that will make both curves have the same common tangent line is
`a = e^(1/e)` which is in decimals a = 1.44467, the tangent line happens at x = e.

Explanation:

Since graphs of the two functions have the same tangent line, means that the slopes will be the same as well as the (x,y) point. Thus we need to find the first derivative to determine the slope of the tangent line and set them equal together to solve for a.

Finding the slope of the tangent lines.

For the first function we have for its derivative

`y = a^x \\ y' =a^x  \ln (a)\\`

For the second function its derivative will be

`y =\log_a (x) \\y'=\cfrac 1{x\ln(a)}`

Thus we can set both derivatives equal to each other since we both should have the same slope of the tangent line.

`a^x \ln (a) =\cfrac 1{x\ln(a)}`

This is our first equation the slope equation.

Setting system of equations

Using the given equations, we can set them equal to each other since we want the point (x,y) to be the same for both lines, so we get

`a^x =\log_a (x)`

We can write it in terms of the natural logarithms using change of base

`a^x = \cfrac{\ln x}{\ln a}`

Solving the system of equations

Replacing the second equation on the first one give us

`\cfrac{\ln x}{\ln a}\ln (a) =\cfrac 1{x\ln(a)}`

Simplifying.

`\ln x =\cfrac 1{x\ln(a)}`

Solving for a

`\ln a =\cfrac 1{x\ln(x)}`

Raising both sides to e^

`a = e^{\cfrac{1}{x\ln x}}`

Here we can pick any value of x except 1 to make both curves to make a tangent line, we can use arbitrarily x = e so we get

`a = e^{\cfrac{1}{e\ln e}}\\a = e^{\cfrac{1}{e}}`

Thus we can conclude that we have a common tangent line at x = e with the value a = 1.44467 or
`a = e^{\cfrac{1}{e}}`.