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

Dec 11, 2016

We will need to differentiate the function

$y = {\log}_{10} \left(2 x\right)$

${10}^{y} = 2 x$

$\ln \left({10}^{y}\right) = \ln \left(2 x\right)$

yln10 = ln(2x)#

$y = \ln \frac{2 x}{\ln} 10$

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

$\left(\ln 2 x\right) ' = \frac{1}{2 x} \times 2 = \frac{2}{2 x} = \frac{1}{x}$

$y ' = \frac{\frac{1}{x} \times \ln 10 - \ln \left(2 x\right) \times 0}{\ln 10} ^ 2$

$y ' = \frac{\ln \frac{10}{x}}{{\ln}^{2} 10}$

$y ' = \ln \frac{10}{x {\ln}^{2} 10}$

$y ' = \frac{1}{x \ln 10}$

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

${m}_{\text{tangent}} = \frac{1}{5 \ln 10}$

${m}_{\text{tangent}} = \frac{1}{\ln} 100000$

We now find the equation:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 1 = \frac{1}{\ln} 100000 \left(x - 5\right)$

$y - 1 = \frac{1}{\ln} 100000 x - \frac{5}{\ln} 100000$

$y = \frac{1}{\ln} 100000 x - \frac{5}{\ln} 100000 + 1$

For an approximation:

$y = 0.08686 x + 0.5657$

Hopefully this helps!