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My son has the following problem he needs help with:

Find a possible formula for the exponential function with the following two points: $(0, 30)$ and $(6, 5)$. Equation Form = f(x)=a*b^x

Can somebody tell me how to do it so I can teach him?

Thanks

NathanMNdz2
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  • You need to specifically define what set of functions you consider exponential. Is it $f(x)=a^x$ or $f(x)=a\cdot b^x$ or what? You likely need two parameters to fit two points. Plug the points into your form and you get two simultaneous equations. Solve them. – Ross Millikan Jan 18 '19 at 16:50

2 Answers2

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Assume $f(x)=Ae^x+B$ is such a function. Then we know $f(0)=30$ and $f(6)=5$. However, $f(0)=Ae^0+B=A+B$ and $f(6)=Ae^6+B$. This implies the following system of equations holds: $$ \begin{cases} A+B=30 \\ Ae^6+B=5 \end{cases} $$ Subtracting the equations gives $(A+B)-(Ae^6+B)=30-5$ or $A-Ae^6=25$ meaning $A(1-e^6)=25$ or $A=\frac{25}{1-e^6}$. From the first equation, we deduce $B=30-\frac{25}{1-e^6}$. We conclude $f(x)=\frac{25}{1-e^6}e^x+30-\frac{25}{1-e^6}=\frac{25}{1-e^6}(e^x-1)+30$ is such a function.

Yuval Gat
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It depends on what you mean by "exponential". I'll assume you mean $f(x)=Ae^{Bx}$ with parameters $A,B$. As a good rule of thumb, you need two equations to get two unknowns and here you have two equations.

We get $$30=f(0)=Ae^0=A\implies A=30$$

and $$5=f(6)=30e^{6B}\implies e^{6B}=\frac 16\implies B=-\frac {\ln 6}6$$

lulu
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