Question

How do you find the equation of the tangent line to the graph $y = {\log _{10}}\left( {2x} \right)$ through point $(5,1)$ ?

Answer 1

Hint : Here we first need to find the slope of the equation that is we differentiate it $\dfrac{{dy}}{{dx}}$ . After that we know that the equation of a tangent is $y - {y_1} = m\left( {x - {x_1}} \right)$ , where ‘m’ is the slope and $\left( {{x_1},{y_1}} \right)$ is the point where tangent line passes through.

Complete step-by-step answer :
Given, $y = {\log _{10}}\left( {2x} \right)$
We need this in natural logarithm form.
The given $y = {\log _{10}}\left( {2x} \right)$ can be written as,
${10^y} = \left( {2x} \right)$
Applying natural log on both sides we have,
$\ln {10^y} = \ln \left( {2x} \right)$
$y\ln 10 = \ln \left( {2x} \right)$
$\Rightarrow y = \dfrac{{\ln \left( {2x} \right)}}{{\ln \left( {10} \right)}}$
Differentiate with respect to ‘x’.
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{\ln \left( {2x} \right)}}{{\ln \left( {10} \right)}}} \right)$
Since $\ln 10$ is constant we can take it outside.
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\ln \left( {10} \right)}}\dfrac{d}{{dx}}\left( {\ln \left( {2x} \right)} \right)$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\ln \left( {10} \right)}}\dfrac{2}{{2x}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{x\ln \left( {10} \right)}}$
But we have $x = 5$ substituting we have,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{5\ln \left( {10} \right)}}$
Slope (m) of a tangent curve is $m = \dfrac{{dy}}{{dx}}$
Hence, $m = \dfrac{1}{{5\ln \left( {10} \right)}}$ .
We also have $\left( {{x_1},{y_1}} \right) = \left( {5,1} \right)$ .
We know the equation of tangent is
$y - {y_1} = m\left( {x - {x_1}} \right)$
Substituting we have,
$y - 1 = \dfrac{1}{{5\ln \left( {10} \right)}}\left( {x - 5} \right)$
$y - 1 = \dfrac{x}{{5\ln \left( {10} \right)}} - \dfrac{5}{{5\ln \left( {10} \right)}}$
$y - 1 = \dfrac{x}{{5\ln \left( {10} \right)}} - \dfrac{1}{{\ln \left( {10} \right)}}$
$\Rightarrow y - \dfrac{x}{{5\ln \left( {10} \right)}} = 1 - \dfrac{1}{{\ln \left( {10} \right)}}$ . This is the required equation.
So, the correct answer is “ $y - \dfrac{x}{{5\ln \left( {10} \right)}} = 1 - \dfrac{1}{{\ln \left( {10} \right)}}$”.

Note : If we have logarithm base other than 10 we use the formula ${\log _a}b = \dfrac{{\ln b}}{{\ln x}}$ to convert it into natural logarithm. We need to know logarithm laws to simplify this. That is the logarithm power rule $\log {\left( a \right)^b} = b\log \left( a \right)$ . Logarithm product law, $\log \left( {ab} \right) = \log \left( a \right) + \log \left( b \right)$ and logarithm quotient rule $\log \left( {\dfrac{a}{b}} \right) = \log \left( a \right) - \log \left( b \right)$ . In all the above laws the bases are the same that is base 10.