How do you find the equation of the tangent line to the graph y=log_10(2x)y=log10(2x) through point (5,1)?
1 Answer
Dec 11, 2016
We will need to differentiate the function
10^y = 2x10y=2x
ln(10^y) = ln(2x)ln(10y)=ln(2x) yln10 = ln(2x)#
y = ln(2x)/ln10y=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
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!
