If f, of, xf(x) is an exponential function where f, of, 1, point, 5, equals, 7f(1.5)=7 and f, of, 8, point, 5, equals, 46f(8.5)=46, then find the value of f, of, 14f(14), to the nearest hundredth.
We are given that $f(1.5)=7$ and $f(8.5)=46$. Let $x$ be an arbitrary real number, and let $xf(x)=y$. This equation can be rewritten as \[f(x)=\frac{y}{x}.\] In particular, we can consider the equation when $x=14$, so $y=14f(14)$. Since this function is exponential, we hypothesize that the given points lie on an exponential curve. We can therefore consider two useful points.
Let us examine the ratio $\frac{f(8.5)}{f(1.5)}$ to see if we can deduce its value. This ratio can be computed as \[\frac{f(8.5)}{f(1.5)}=\frac{14\cdot f(8.5)}{14\cdot f(1.5)}=\frac{14f(8.5)}{14f(1.5)}=\frac{14f(8.5)}{7}=\frac{46}{7}=\frac{46}{7}.\] Since we see that this ratio is equal to $\frac{46}{7}$, we can guess that the function $f(x)$ is in fact such that $f(x)=ae^{bx}$ for some constants $a$ and $b$.
Proceeding with this information, we know that $f(1.5)=ae^{1.5b}=7$. Similarly, we know that $f(8.5)=ae^{8.5b}=46$. Dividing these two equations, we get \[\frac{f(8.5)}{f(1.5)}=\frac{ae^{8.5b}}{ae^{1.5b}}=e^{7b},\] where we may divide out the constant $a$. Thus, we have that $e^{7b}=\frac{46}{7}$. Taking the natural logarithm of both sides, we derive $7b=\ln\left(\frac{46}{7}\right)$. We can then divide both sides by 7 to solve for $b$: \[b=\frac{\ln\left(\frac{46}{7}\right)}{7}=\frac{\ln 46-\ln 7}{7}.\]
Now we may substitute $b$ into the original equations. We have that $ae^{1.5b}=7$, so $ae^{1.5\cdot\frac{\ln 46-\ln 7}{7}}=7$. Dividing by $e^{\frac{1.5\cdot\ln 46-1.5\cdot\ln 7}{7}}=e^{\ln\sqrt{46}-\ln\sqrt{7}}$, we get that \[a=e^{-\frac{\ln\sqrt{46}-\ln\sqrt{7}}{7}}=e^{\frac{\ln\sqrt{7}-\ln\sqrt{46}}{7}}=e^{\frac{\ln \frac{\sqrt{7}}{\sqrt{46}}}{7}}.\] Writing this in simplified form gives \[a=e^{\frac{\ln \frac{\sqrt{\frac{7}{46}}}}{7}}=\left(e^{\ln \frac{\sqrt{\frac{7}{46}}}}\right)^{\frac{1}{7}}=\left(\frac{\sqrt{\frac{7}{46}}}{e^{\ln 2}}\right)^{\frac{1}{7}}=\left(\frac{\sqrt{\frac{7}{46}}}{2}\right)^{\frac{1}{7}}=\left(\frac{\sqrt{7}}{2\sqrt{46}}\right)^{\frac{1}{7}}.\]
To find $14f(14)$ to the nearest hundredth, we let $x=14$, so \[y=14f(14)=14\cdot(ae^{14b})=14\cdot\left(\left(\frac{\sqrt{7}}{2\sqrt{46}}\right)^{\frac{1}{7}}e^{14\cdot\frac{\ln\sqrt{46}-\ln\sqrt{7}}{7}}\right).\] This simplifies as \[14\left(\frac{\sqrt{7}}{2\sqrt{46}}\right)^{\frac{1}{7}}\left(e^{\ln 46-\ln 7}\right)^{2}=14\left(\frac{\sqrt{7}}{2\sqrt{46}}\right)^{\frac{1}{7}}\frac{46^2}{7^2}=14\frac{46^2}{2^2\cdot 7^2\cdot 46^{\frac{6}{7}}}.\] Simplifying further, we have that this expression equals \[14\frac{46^2}{2^2\cdot 7^2\cdot 46^{\frac{6}{7}}}=14\frac{46^2}{2^2\cdot 7^2\cdot 46^{\frac{42}{7}}}=14\frac{46^2}{2^2\cdot 7^2\cdot 46^6}=7\cdot \frac{46^2}{2^2\cdot 7^2\cdot 46^6}=7\cdot\frac{1}{2^2\cdot 46^4}=\frac{46^4}{2^2\cdot 46^4\cdot 2}=7\cdot\frac{1}{2^2}=\boxed{1.75}.\]