Question 430·Hard·Math

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Function $g$ is defined by

g(x)=k\,r^{x}-1,

where $k$ and $r$ are constants with $r>0$ . In the $xy$ -plane, the graph of $y=g(x)$ passes through the points $(1,7)$ and $(3,31)$ . What is the value of $r$ ?

Your answer

A shifted exponential function has a fixed amount added after the power, so its base multiplies only what remains when that amount is removed. Given two points and unknown constants, turn each point into an output equation, then fit both equations with a Desmos list regression. Don't divide the original outputs, and remember that an input gap can cover more than one multiplication.

Hints

Hint 1
A point on a function's graph pairs an input with its output: $(a,b)$ means $g(a)=b$ . What equations do you get by putting the two given inputs into the rule?
Hint 2
The $-1$ is a shift outside the power. Add $1$ to each output to uncover the amounts that the base multiplies.
Hint 3
A regression can find constants that fit several conditions at once. Put the two shifted equations into one Desmos list regression, and include the given restriction $r>0$ .

Step-by-step

Approach 1: Fit the shifted outputs in Desmos
Step 1
Turn the points into equations
A point on $y=g(x)$ tells you an input and its output. So $(1,7)$ means $g(1)=7$ , and $(3,31)$ means $g(3)=31$ :
$\begin{aligned} kr-1&=7,\\[1.4em] kr^3-1&=31. \end{aligned}$
Step 2
Remove the fixed shift
The $-1$ is a fixed shift outside the power. Add $1$ to each output to find the amounts multiplied by $r$ :
$kr=7+1=8,\qquad kr^3=31+1=32.$
Compare the shifted amounts, not the original outputs, to find an exponential base.
Step 3
Fit both equations
Type $[kr,kr^3]\sim[8,32]\{r>0\}$ into Desmos. The tilde tells Desmos to fit both equations; the restriction keeps the base positive as required. Under PARAMETERS, Desmos shows $k=4$ and $r=2$ . So the base is $r=2$ . Grid in 2.

Approach 2: See why the fitted base works
Step 1
Cancel the unknown multiplier
Instead of fitting $k$ , divide the shifted equations. The input rises from $1$ to $3$ , a two-step gap, so the powers leave $r^2$ . Division is safe because $kr=8$ , not zero. Divide the equations:
$\frac{kr^3}{kr}=\frac{32}{8}.$
Cancel $k$ and one $r$ :
$r^2=4.$
Step 2
Choose the allowed base
A square can have a positive or negative root, but the positive-base condition says $r>0$ . Take the positive square root:
$r=\sqrt{4}=2.$
So the exponential base is $r=2$ . Grid in 2.

Question 430·Hard·Math

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