Find the equation of the tangent line to the curve when x is equal to 2. `y=log_2 x+log_2(x+6)`

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`y= log_2 x + log_2(x+6)`

We need to find the equation of the tangent line when x= 2.

First, we will find the y-coordinate when x= 2

`==gt y(2)= log_2 2 + log_2 8 = 1 + 3 = 4`

==> Then we need to find the equation of the tangent line at the point (2,4)

==> `y-y1= m (x-x1)`

==> `y-4 = m(x-2)`

Now we will find the slope.

We know that the slope m is the derivative at x= 2.

==> `y'= 1/(xln2) + 1/(x+6)ln2 `

==>`y'(2)= 1/(2ln2) + 1/(8ln2) = 5/(8ln2) ~~ 0.1803 `

`==gt y-4 = (0.1803) (x-2)`

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