How do you find the equation of the tangent line to the graph $y = {log}_{3} x$ $y=log_3x$ through point (27,3)?

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How do you find the equation of the tangent line to the graph $y = {log}_{3} x$ $y=log_3x$ through point (27,3)?

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Answer 1 · 2017-05-22T07:39:10

$x - 27 ln 3 - 27 + 81 ln 3 = 0$ $x-27ln3-27+81ln3=0$

Explanation:

As slope of tangent to a curve at a given point is give by the value of first derivative at that point, let us first find the differential of $y = {log}_{3} x$ $y=log_3x$ .

As $y = {log}_{3} x = \frac{ln x}{ln 3}$ $y=log_3x=lnx/ln3$

$\frac{d y}{d x} = \frac{1}{x ln 3}$ $(dy)/(dx)=1/(xln3)$ and at $(27, 3)$ $(27,3)$ slope is $\frac{1}{27 ln 3}$ $1/(27ln3)$

Hence equation of tangent is

$y - 3 = \frac{1}{27 ln 3} (x - 27)$ $y-3=1/(27ln3)(x-27)$

or $27 ln 3 y - 81 ln 3 - x + 27 = 0$ $27ln3y-81ln3-x+27=0$

or $x - 27 ln 3 - 27 + 81 ln 3 = 0$ $x-27ln3-27+81ln3=0$

graph{(x-27ln3y-27+81ln3)(y-lnx/ln3)((x-27)^2+(y-3)^2-0.05)=0 [-10, 30, -10, 10]}

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How do you find the equation of the tangent line to the graph $y = {log}_{3} x$ $y=log_3x$ through point (27,3)?

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Explanation:

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How do you find the equation of the tangent line to the graph $y = {log}_{3} x$ $y=log_3x$ through point (27,3)?