New Notes Blog: `(4,3) , (8,15)` Write a power function `y=ax^b` whose graph passes through the given points

Sunday, 9 August 2015

$\displaystyle{\left({4},{3}\right)},{\left({8},{15}\right)}$ Write a power function $\displaystyle{y}={a}{{x}}^{{b}}$ whose graph passes through the given points

We are asked to write the equation for a power function whose graph passes through the points (4,3) and (8,15).

We substitute the known values of x and y into the basic equation to get two equations with two unknowns (a and b) and then solve the system for the coefficients.

$\displaystyle{3}={a}\cdot{{4}}^{{b}},{15}={a}\cdot{{8}}^{{b}}$

Solving the first equation for a we get:

$\displaystyle{a}=\frac{{3}}{{{{4}}^{{b}}}}$

Substitute this expression for a in the second equation to get:

$\displaystyle{15}=\frac{{3}}{{{{4}}^{{b}}}}\cdot{{8}}^{{b}}$

$\displaystyle{15}={3}\cdot{{\left(\frac{{8}}{{4}}\right)}}^{{b}}$

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We are asked to write the equation for a power function whose graph passes through the points (4,3) and (8,15).

We substitute the known values of x and y into the basic equation to get two equations with two unknowns (a and b) and then solve the system for the coefficients.

$\displaystyle{3}={a}\cdot{{4}}^{{b}},{15}={a}\cdot{{8}}^{{b}}$

Solving the first equation for a we get:

$\displaystyle{a}=\frac{{3}}{{{{4}}^{{b}}}}$

Substitute this expression for a in the second equation to get:

$\displaystyle{15}=\frac{{3}}{{{{4}}^{{b}}}}\cdot{{8}}^{{b}}$

$\displaystyle{15}={3}\cdot{{\left(\frac{{8}}{{4}}\right)}}^{{b}}$

$\displaystyle{{2}}^{{b}}={5}$

So $\displaystyle{b}=\frac{{{\ln{{\left({5}\right)}}}}}{{{\ln{{\left({2}\right)}}}}}\approx{2.322}$

Now substitute for b to get a:

$\displaystyle{a}=\frac{{3}}{{{{4}}^{{b}}}}=\frac{{3}}{{{{4}}^{{\frac{{{\ln{{\left({5}\right)}}}}}{{{\ln{{\left({2}\right)}}}}}}}}}=\frac{{3}}{{25}}={.12}$

So the model is $\displaystyle{y}=\frac{{3}}{{25}}{{x}}^{{\frac{{{\ln{{\left({5}\right)}}}}}{{{\ln{{\left({2}\right)}}}}}}}\text{or}{y}\approx{.12}{{x}}^{{2.322}}$