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

Apr 20, 2017

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:

$\ln \left(y\right) = \ln \left({2}^{-} x\right)$

Use the property $\ln \left({a}^{c}\right) = \left(c\right) \ln \left(a\right)$

$\ln \left(y\right) = - x \ln \left(2\right) = \ln \left(\frac{1}{2}\right) x$

Differentiate both sides:

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \ln \left(\frac{1}{2}\right)$

Multiply both sides by y:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \ln \left(\frac{1}{2}\right) y$

Substitute for y:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \ln \left(\frac{1}{2}\right) {2}^{-} x$

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

$m = \ln \left(\frac{1}{2}\right) {2}^{-} \left(- 1\right)$

$m = 2 \ln \left(\frac{1}{2}\right)$

$m = \ln \left(\frac{1}{4}\right)$

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

$y = m \left(x - {x}_{1}\right) + {y}_{1}$

$y = \ln \left(\frac{1}{4}\right) \left(x - - 1\right) + 2$

$y = \ln \left(\frac{1}{4}\right) \left(x + 1\right) + 2$