Question

What is the value of `x` where the tangent to `y -1 = 3^x` has a slope of `5`?

Answer 1

`x= ln(5/ln3)/ln3 ~=1.379`

Explanation:

Start by finding the derivative of the function.

`y - 1 = 3^x`

`ln(y - 1) = ln(3^x)`

`ln(y - 1) = xln3`

Differentiate the left hand side using the chain rule and the right hand side using the product rule.

`1/(y - 1) xx1(dy/dx) = 1(ln3) + 0(x)`

`1/(y- 1)dy/dx= ln3`

`dy/dx = ln3/(1/(y - 1))`

`dy/dx= ln3(y - 1)`

`dy/dx= ln3(3^x + 1 - 1)`

`dy/dx= 3^xln3`

The derivative represents the instantaneous rate of change of a function at any given point `x= a` within the domain of the function. In this problem, we want to find where the tangent line is parallel to `y = 5x- 1`. The line `y =5x - 1` has a slope of `5`, parallel lines have equal slopes, so we will want to find where on the derivative the `dy/dx = 5`.

`5 = 3^xln3`

`5/ln3 = 3^x`

`ln(5/ln3) = xln3`

`ln(5/ln3)/ln3 = x`

If you would like an approximation, use a calculator to obtain `x = 1.379`.

Hopefully this helps!