I am asked to find an exponential equation that passes through $(2, 2.25)$ and $(5, 60.75)$. My textbook says the solution it's $y=0.25(3)^x$ whereas I got $y=0.028(9)^x$. Here is my working:
Express each coordinate in terms of the exponential function:$$2.25=ab^2$$$$60.75=ab^5$$
Solve for $a$ in terms of $b$ in the first equation:$$a=\frac{2.25}{b^2}$$$$a=2.25b^{-2}$$
Substitute $a$ in the second equation and solve for $b$:$$60.75=2.25b^{-2}b^5$$$$60.75=2.25b^3$$$$b^3=\frac{60.75}{2.25}$$$$b=27^{\frac{1}{3}}$$$$b=9$$
Use the value of b in the first equation to solve for the value of a:$$a=2.25b^{-2}$$$$a=2.25(9)^{-2}$$$$a=\frac{2.25}{9^2}$$$$a=0.028$$
Thus my solution is:$$y=0.028(9)^x$$
Where did I go wrong and how can I arrive at $y=0.25(3)^x$?
$\endgroup$ 13 Answers
$\begingroup$You have mistakenly wrote:$$b = 27^{\frac{1}{3}}$$$$ b = 9 $$
When in fact $b = 27^{\frac{1}{3}} = 3$.
$\endgroup$ 2 $\begingroup$I think the easiest way to do it would be to divide one equation by the other then sub back in:$$\frac{ab^5}{ab^2}=\frac{60.75}{2.25}$$$$b=\left(\frac{60.75}{2.25}\right)^{1/3}=\sqrt[3]{27}=3$$wheres you said this value is equal to $9$, so that is your mistake
$\endgroup$ 2 $\begingroup$Alternatively, taking logarithms on $y = ab^x$, $\ln y = \ln a + x \ln b$, which is a linear equation with $\ln y$ against $x$, as you can rewrite this as $y' = c + mx$.
The line passing through $(2, \ln 2.25)$ and $(5, \ln 60.75)$ has gradient $\frac{\ln 60.75 - \ln 2.25}{5 - 2} = \frac{\ln 27}{3}$, so $\ln b = \frac{\ln 27}{3}, b = e^{(\ln 27) / 3} = 27^{1/3} = 3$. Also, $\ln a = \ln y - x \ln b$, where using the point $(2, \ln 2.25)$ gives $\ln a = \ln 2.25 - (2 \ln 27) / 3 = \ln 2.25 - \ln 27^{2/3} = \ln 0.25$, thus $a = 0.25$. So since $a = 0.25, b = 3$, the equation passing through these two points is $y = 0.25 \cdot 3^x$.
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