Question 856·Hard·Math

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A function can be written in the form $f(x) = a b^x + c$ , where $a$ , $b$ , and $c$ are real constants with $a \ne 0$ and $b > 1$ . Suppose $f(1) - f(0) = 12$ and $f(3) - f(2) = 108$ . What is the value of $b$ ?

The cue is a shifted exponential $ab^x+c$ with differences between consecutive outputs. Subtracting outputs removes the fixed $c$ , and each one-step difference grows by a factor of $b$ . Compare how far apart the two given differences are, then solve for the base. Take a ratio of changes, not a ratio of outputs; the shift would spoil an output ratio.

Hints

Hint 1
An output difference subtracts one function value from another. Because $c$ is added to both values, it cancels. What do you get when you substitute the rule into $f(x+1)-f(x)$ ?
Hint 2
Each time the input increases by $1$ , a one-step difference is multiplied by $b$ . The given differences begin at inputs $0$ and $2$ , so they're two steps apart. What power of $b$ links them?
Hint 3
Divide the later difference by the earlier one to cancel the shared factor $a(b-1)$ . The result is a power of $b$ , not necessarily $b$ itself. How does $b>1$ help you pick the right root?

Step-by-step

Compare the two output changes
Step 1
Find the change across one input step
The $c$ in $f(x)=ab^x+c$ is a fixed shift: it's added to every output, so it disappears when you subtract two outputs. Substitute $x+1$ and $x$ into the rule:
$f(x+1)-f(x)=(ab^{x+1}+c)-(ab^x+c).$
Cancel $c$ :
$f(x+1)-f(x)=ab^{x+1}-ab^x.$
Use $b^{x+1}=b^xb$ :
$f(x+1)-f(x)=ab^xb-ab^x.$
Factor out $ab^x$ :
$f(x+1)-f(x)=ab^x(b-1).$
Step 2
Match the formula to both given changes
The first change starts at $x=0$ . Use $b^0=1$ to match it to the given $12$ :
$12=a(b-1).$
The second change starts at $x=2$ , so match it to the given $108$ :
$108=ab^2(b-1).$
Step 3
Divide the changes to isolate the base
Because $a\ne0$ and $b>1$ , the shared factor $a(b-1)$ isn't zero. Divide the second equation by the first, so that factor cancels:
$\frac{108}{12}=\frac{ab^2(b-1)}{a(b-1)}.$
Cancel it:
$\frac{108}{12}=b^2.$
Divide $108$ by $12$ :
$9=b^2.$
The $9$ is $b$ squared, not $b$ .
Step 4
Find the allowed base
Type $b_1^2\sim9\{b_1>1\}$ in Desmos. The subscript makes $b_1$ a value Desmos can fit, $\sim$ asks it to solve the equation, and the restriction keeps only the root greater than $1$ . Desmos shows $b_1=3$ under PARAMETERS. So the base is $b=3$ . Choice D.

Question 856·Hard·Math

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