Question

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

Answer 1

Use logarithmic differentiation.
Evaluate the derivative at `x = -1`, to find the slope.
Use the point-slope form of the equation of the line.

Explanation:

Use the natural logarithm on both sides of the equation:

`ln(y)=ln(2^-x)`

Use the property `ln(a^c) = (c)ln(a)`

`ln(y)=-xln(2) = ln(1/2)x`

Differentiate both sides:

`1/ydy/dx=ln(1/2)`

Multiply both sides by y:

`dy/dx=ln(1/2)y`

Substitute for y:

`dy/dx=ln(1/2)2^-x`

The slope, m, of the tangent line is the derivative evaluated at `x=-1`:

`m=ln(1/2)2^-(-1)`

`m=2ln(1/2)`

`m=ln(1/4)`

Using the point-slope form of the equation of the line:

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

`y = ln(1/4)(x--1)+2`

`y = ln(1/4)(x+1)+2`