Question

How do you find the equation of the tangent line to the graph `y=5^(x-2)` through point (2,1)?

Answer 1

Find the derivative of the function.

`y= 5^(x- 2)`

`lny = ln(5^(x- 2))`

`lny = (x- 2)ln5`

`1/y(dy/dx) = 1(ln5) + (x- 2)0`

`1/y(dy/dx) = ln5`

`dy/dx = ln5/(1/y)`

`dy/dx= ln5(5^(x - 2))`

The slope of the tangent is therefore `ln5(5^(2- 2)) = ln5`.

`y -y_1 = m(x- x_1)`

`y - 1 = ln5(x- 2)`

`y - 1= ln5x - 2ln5`

`y = ln5x - 2ln5 + 1`

Hopefully this helps!