Exponential function - with one point in fraction form

Question

Question: determine the exponential function when given 2 points: (-1,2) and (3/2,1/2). The examples I've found are with integers, not fractions, and I'm not sure how/if that effects the answer...

Answer 1

One way to do this problem is using the formula $y=ab^x$. Now if $y=2$ then $x=-1$ so that gives $2=ab^{-1}$. And also if $y=0.5$ then $x=1.5$ which gives $0.5=ab^{1.5}$. When you divide first equation by second, the $a$ cancels and you are left with $4=b^{-2.5}$ from which you can find $b$. This can be written as $\frac{1}{b^{2.5}}=4$ or $b^{2.5}=\frac{1}{4}$. Raising both sides to the power $2.5$ gives $b=\frac{1}{32}$. You can now backsub to find $a$

Answer 2

More generally, if $ab^x$ passes through $(p, q)$ and $(r, s)$, then $ab^p = q$ and $ab^r = s$.

Dividing these, $b^{p-r} = \frac{q}{s}$ or $b =(q/s)^{1/(p-r)} $.

Then $a =qb^{-p} =q(q/s)^{-p/(p-r)} =q^{1-p/(p-r)}s^{p/(p-r)} =q^{-r/(p-r)}s^{p/(p-r)} $.

In your case, $p, q, r, s =-1, 2, 3/2, 1/2 $.

Then $1/(p-r) =1/(-1-3/2) =-1/(5/2) =-2/5 $ so $b =(2/(1/2))^{-2/5} =4^{-2/5} =2^{-4/5} $ and $a =qb^{-p} =2(2^{-4/5})^{1} =2(2^{-4/5}) =2^{1/5} $.