Question

How do you find the equation of the tangent line to the graph `y=log_10(2x)` through point (5,1)?

Answer 1

We will need to differentiate the function

`y = log_10(2x)`

`10^y = 2x`

`ln(10^y) = ln(2x)`

yln10 = ln(2x)#

`y = ln(2x)/ln10`

We differentiate the numerator of this expression using the chain rule and the entire function using the quotient rule.

`(ln2x)' = 1/(2x) xx 2 = 2/(2x) = 1/x`

`y' = (1/x xx ln10 - ln(2x) xx 0)/(ln10)^2`

`y' = (ln10/x)/(ln^2 10)`

`y' = ln10/(xln^2 10)`

`y' = 1/(xln10)`

The slope of the tangent is given by substituting `x = a` into the derivative.

`m_"tangent" = 1/(5ln10)`

`m_"tangent" = 1/ln100000`

We now find the equation:

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

`y - 1 = 1/ln100000(x - 5)`

`y - 1 = 1/ln100000x - 5/ln100000`

`y = 1/ln100000x - 5/ln100000 + 1`

For an approximation:

`y = 0.08686x + 0.5657`

Hopefully this helps!