Question 17·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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For a positive constant $k$ , the graphs of the equations $y=3^{x+1}+k$ and $y=9^x+2k$ intersect at two distinct points. The $x$ -coordinates of the intersection points differ by $2$ . Which choice gives the value of $k$ ?

Two exponential graphs form a system of equations at their intersections: both equations must give the same $y$ -value. Rewriting $9^x$ as $(3^x)^2$ turns that shared value into an expression involving one power. Convert the gap between the intersection inputs into a multiplicative change in that power, then use a Desmos regression to find it. Don’t mistake the power for the constant the question asks for.

Hints

Hint 1
At an intersection, the two graphs have the same output. Set their expressions for $y$ equal, then move the $k$ terms together. What expression for $k$ depends only on $x$ ?
Hint 2
Rewrite $9^x$ as $(3^x)^2$ . If $t=3^x$ , the expression for $k$ becomes a quadratic, an expression with a $t^2$ term. Keep that squared term.
Hint 3
Increasing an exponent by $2$ multiplies $3^x$ by $3^2$ ; it doesn’t add $2$ to the power. How much larger is the power at the second intersection? Both intersections must give the same $k$ .

Step-by-step

Match the constant at both intersections
Step 1
Express the constant using an intersection
At an intersection, the graphs have the same $y$ -value, so set their expressions for $y$ equal:
$3^{x+1}+k=9^x+2k.$
Subtract $k$ from both sides:
$3^{x+1}=9^x+k.$
Subtract $9^x$ from both sides:
$k=3^{x+1}-9^x.$
Step 2
Write the constant in terms of one power
Let $x$ be the smaller intersection’s $x$ -coordinate, and set $t=3^x$ . Since $9=3^2$ , use the power-of-a-power rule:
$9^x=(3^x)^2=t^2.$
Also, $3^{x+1}=3(3^x)=3t$ . So $k=3t-t^2$ . Type $f(t)=3t-t^2$ in Desmos; this function gives the value of $k$ for an intersection whose power is $t$ .
Step 3
Use the gap between the two inputs
The other intersection is at $x+2$ . An exponent increase of $2$ multiplies the power by $9$ :
$3^{x+2}=3^x\cdot3^2=9t.$
Both intersections give the same fixed $k$ , so $f(t)=f(9t)$ . Adding $2$ to $t$ instead would confuse a change in the exponent with a change in the power.
Step 4
Find the smaller positive power
Type $f(t_1)\sim f(9t_1)\{t_1>0\}$ below the function. The $\sim$ asks Desmos to fit an equality, and $t_1$ is the number it finds. The positive restriction matters: $t=3^x$ can’t be $0$ , even though $0$ also makes the two expressions equal. Desmos shows $t_1=0.3$ under PARAMETERS.
Step 5
Evaluate the constant
Type $f(t_1)$ on the next line. Desmos shows $0.81$ , which is $\frac{81}{100}$ . That’s $k$ , not the power $t_1$ . So the positive constant that makes the intersection $x$ -coordinates differ by $2$ is $k=\frac{81}{100}$ . Choice A.

Question 17·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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