`y = log_10(2x) , (5,1)` Find an equation of the tangent line to the graph of the function at the given point

`y = log_10 (2x)`

The line is tangent to the graph of function at (5,1).  The equation of the tangent line is _______. 

To solve this, we have to determine the slope of the tangent. Take note that the slope of a tangent is equal to the derivative of the function.

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

Applying this,  the y' will be:

`y' =...

`y = log_10 (2x)`

The line is tangent to the graph of function at (5,1).  The equation of the tangent line is _______. 

To solve this, we have to determine the slope of the tangent. Take note that the slope of a tangent is equal to the derivative of the function.

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

Applying this,  the y' will be:

`y' = d/dx [log_10 (2x)]`

`y' = 1/(ln(10) * 2x) * d/dx (2x)`

`y'=1/(ln(10) * 2x) *2`

`y'=1/(xln(10))`

Then, plug-in the given point of tangency to get the slope.  

`y'= 1/(5ln(10))`

So the line that is tangent to the graph of the function at point (5,1) has a slope of `m=1/(5ln(10))` .

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

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

Plugging in the values, it becomes:

`y - 1 = 1/(5ln(10))(x - 5)`

`y - 1 = 1/(5ln(10))*x - 1/(5ln(10))*5`

`y - 1 =x/(5ln(10)) -1/(ln(10))`

`y=x/(5ln(10)) - 1/(ln(10))+1`

Therefore, the equation of the tangent line is  `y=x/(5ln(10)) - 1/(ln(10))+1` .