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

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

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Answer 1 · 2016-12-13T20:50:50

Find the derivative of the function.

$y = 5^{x - 2}$ $y= 5^(x- 2)$

$ln y = ln (5^{x - 2})$ $lny = ln(5^(x- 2))$

$ln y = (x - 2) ln 5$ $lny = (x- 2)ln5$

$\frac{1}{y} (\frac{d y}{d x}) = 1 (ln 5) + (x - 2) 0$ $1/y(dy/dx) = 1(ln5) + (x- 2)0$

$\frac{1}{y} (\frac{d y}{d x}) = ln 5$ $1/y(dy/dx) = ln5$

$\frac{d y}{d x} = \frac{ln 5}{\frac{1}{y}}$ $dy/dx = ln5/(1/y)$

$\frac{d y}{d x} = ln 5 (5^{x - 2})$ $dy/dx= ln5(5^(x - 2))$

The slope of the tangent is therefore $ln 5 (5^{2 - 2}) = ln 5$ $ln5(5^(2- 2)) = ln5$ .

$y - y_{1} = m (x - x_{1})$ $y -y_1 = m(x- x_1)$

$y - 1 = ln 5 (x - 2)$ $y - 1 = ln5(x- 2)$

$y - 1 = ln 5 x - 2 ln 5$ $y - 1= ln5x - 2ln5$

$y = ln 5 x - 2 ln 5 + 1$ $y = ln5x - 2ln5 + 1$

Hopefully this helps!

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

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