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

Question

Ezekiel Alexander · Accepted Answer

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!

William Abdalla · Answer

undefined To find the equation of the tangent line to the graph $y = \log_{10}(2x)$ through point $(5,1)$, we first find the derivative of the function. Then, we evaluate the derivative at $x = 5$ to find the slope of the tangent line. Finally, we use the point-slope form of a linear equation to write the equation of the tangent line. The derivative of $y = \log_{10}(2x)$ is $\frac{1}{x \ln(10)}$, and evaluating it at $x = 5$ gives us $\frac{1}{5 \ln(10)}$. Using the point-slope form with slope $\frac{1}{5 \ln(10)}$ and point $(5,1)$, the equation of the tangent line is $y - 1 = \frac{1}{5 \ln(10)}(x - 5)$.