# How do you find the equation of the tangent line to the graph #y=log_3x# through point (27,3)?

Hence equation of tangent is

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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To find the equation of the tangent line to the graph ( y = \log_3 x ) through the point ((27, 3)), we first find the derivative of the function. Then we evaluate the derivative at ( x = 27 ) to find the slope of the tangent line. Finally, we use the point-slope form of a linear equation to find the equation of the tangent line.

The derivative of ( y = \log_3 x ) is ( \frac{dy}{dx} = \frac{1}{x \ln(3)} ).

Evaluating the derivative at ( x = 27 ), we have ( \frac{dy}{dx} \bigg|_{x=27} = \frac{1}{27 \ln(3)} ).

This gives us the slope of the tangent line.

Using the point-slope form with the point ((27, 3)) and the slope we found, the equation of the tangent line is ( y - 3 = \frac{1}{27 \ln(3)}(x - 27) ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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