`y = log_3(x) , (27,3)` Find an equation of the tangent line to the graph of the function at the given point

Wednesday, May 11, 2011

$\displaystyle{y}={\log}_{{3}}{\left({x}\right)},{\left({27},{3}\right)}$ Find an equation of the tangent line to the graph of the function at the given point

$\displaystyle{y}={\log}_{{3}}{\left({x}\right)}$

The line is tangent to the graph of the function at (27,3). The equation of the tangent line is _____.

To solve, the slope of the tangent line should be determined. Take note that the slope of a tangent line is equal to the value of the derivative at the point of tangency.

To determine the derivative of the function, apply the formula $\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left[{\log}_{{a}}{\left({u}\right)}\right]}=\frac{{1}}{{{\ln{{\left({a}\right)}}}\cdot{u}}}\cdot\frac{{{d}{u}}}{{\left.{d}{x}}}$ .

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{d}}{{\left.{d}{x}}}{\left[{\log}_{{3}}{\left({x}\right)}\right]}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{1}}{{{\ln{{\left({3}\right)}}}\cdot{x}}}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({x}\right)}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{1}}{{{\ln{{\left({3}\right)}}}\cdot{x}}}\cdot{1}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{1}}{{{x}{\ln{{\left({3}\right)}}}}}$

The point of tangency is (27,3). So plug-in x = 27 to the derivative to get the slope of the tangent.

$\displaystyle{m}=\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{1}}{{{27}{\ln{{\left({3}\right)}}}}}$

Hence, the line that is tangent to the graph of the function at point (27,3) has a slope of $\displaystyle{m}=\frac{{1}}{{{27}{\ln{{\left({3}\right)}}}}}$ .

To get the equation of the line, apply the point-slope form.

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

Plugging in the values of m, x1 and y1, it becomes:

$\displaystyle{y}-{3}=\frac{{1}}{{{27}{\ln{{\left({3}\right)}}}}}{\left({x}-{27}\right)}$

$\displaystyle{y}-{3}=\frac{{1}}{{{27}{\ln{{\left({3}\right)}}}}}\cdot{x}-{27}\cdot\frac{{1}}{{{27}{\ln{{\left({3}\right)}}}}}$

$\displaystyle{y}-{3}=\frac{{x}}{{{27}{\ln{{\left({3}\right)}}}}}-\frac{{1}}{{\ln{{\left({3}\right)}}}}$

$\displaystyle{y}=\frac{{x}}{{{27}{\ln{{\left({3}\right)}}}}}-\frac{{1}}{{\ln{{\left({3}\right)}}}}+{3}$

Therefore, the equation of the tangent line is $\displaystyle{y}=\frac{{x}}{{{27}{\ln{{\left({3}\right)}}}}}-\frac{{1}}{{\ln{{\left({3}\right)}}}}+{3}$ .

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Wednesday, May 11, 2011

$\displaystyle{y}={\log}_{{3}}{\left({x}\right)},{\left({27},{3}\right)}$ Find an equation of the tangent line to the graph of the function at the given point

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Wednesday, May 11, 2011

Find an equation of the tangent line to the graph of the function at the given point

No comments:

Post a Comment

Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle{y}={\log}_{{3}}{\left({x}\right)},{\left({27},{3}\right)}$ Find an equation of the tangent line to the graph of the function at the given point

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?