`y = log_3(x) , (27,3)` Find an equation of the tangent line to the graph of the function at the given point

`y=log_3(x)`

The line is tangent to the graph of the function at (27,3). The equation of the tangent line is _____.

To solve, the slope of the tangent line should be determined. Take note that the slope of a tangent line is equal to the value of the derivative at the point of tangency.

To determine the derivative of the function, apply the formula `d/dx[log_a(u)]=1/(ln(a)*u)*(du)/dx`.

`(dy)/dx = d/dx[log_3 (x)]`

`(dy)/dx =1/(ln(3)*x) * d/(dx)(x)`

`(dy)/dx =1/(ln(3)*x)*1`

...

`y=log_3(x)`

The line is tangent to the graph of the function at (27,3). The equation of the tangent line is _____.

To solve, the slope of the tangent line should be determined. Take note that the slope of a tangent line is equal to the value of the derivative at the point of tangency.

To determine the derivative of the function, apply the formula `d/dx[log_a(u)]=1/(ln(a)*u)*(du)/dx`.

`(dy)/dx = d/dx[log_3 (x)]`

`(dy)/dx =1/(ln(3)*x) * d/(dx)(x)`

`(dy)/dx =1/(ln(3)*x)*1`

`(dy)/dx = 1/(xln(3))`

The point of tangency is (27,3). So plug-in x = 27 to the derivative to get the slope of the tangent.

`m=(dy)/dx = 1/(27ln(3))`

Hence, the line that is tangent to the graph of the function at point (27,3) has a slope of `m = 1/(27ln(3))` .

To get the equation of the line, apply the point-slope form.

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

Plugging in the values of m, x1 and y1, it becomes:

`y - 3= 1/(27ln(3))(x - 27)`

`y-3=1/(27ln(3))*x - 27 * 1/(27ln(3))`

`y - 3 =x/(27ln(3)) - 1/ln(3)`

`y =x/(27ln(3)) - 1/ln(3)+3`

Therefore, the equation of the tangent line is `y =x/(27ln(3)) - 1/ln(3)+3` .