Question 165·200 Super-Hard SAT Math Questions·Advanced Math

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The function $f$ is defined by $f(x)=240(b)^{x/2}$ , where $b>1$ .

When $x$ increases by 6, $f(x)$ increases by $p$ %. When $x$ increases by 5, $f(x)$ increases by $c$ %. Which choice expresses $c$ in terms of $p$ ?

In an exponential model, an increase in the input multiplies the output rather than adding a fixed amount. When the base is unknown but the answers are expressions, you can choose an allowed base and test the choices in Desmos. Convert each percent increase to a multiplier first. To change intervals, raise the old multiplier to the new interval divided by the old interval. Reversing that fraction is the main trap.

Hints

Hint 1
The only restriction on the base is $b>1$ , so you can choose a convenient value such as $4$ . A correct expression for $c$ must work for that value. What percent increases does the function give?
Hint 2
A growth factor is the new output divided by the old output. To find the percent increase over 6 input units, compare $f(6)$ with $f(0)$ , subtract $1$ from the ratio, and multiply by $100$ .
Hint 3
The 5-unit interval is shorter than the 6-unit interval, so its growth factor should be smaller. Find its percent increase the same way, then substitute your value of $p$ into the choices.

Step-by-step

Approach 1: Test an allowed base in Desmos
Step 1
Choose a base that makes the powers manageable
The base $b$ is allowed to be any number greater than $1$ , so choose $b=4$ . Type $b=4$ , then enter the given function as $f(x)=240b^{x/2}$ . You aren't claiming that the unknown base must be $4$ : a correct choice must work for this allowed example.
Step 2
Find the 6-unit percent increase
The ratio $f(6)/f(0)$ is the growth factor from $x=0$ to $x=6$ . The increase is the part above $1$ , so type $p=100(f(6)/f(0)-1)$ . Desmos shows $p=6300$ for the base you chose.
Step 3
Find the 5-unit percent increase
Now compare $f(5)$ with the same starting output, $f(0)$ . Type $c=100(f(5)/f(0)-1)$ . Desmos shows $c=3100$ for this example. Comparing both changes with $f(0)$ keeps the test consistent.
Step 4
Compare the choices with the test result
The choices use powers $\frac12$ , $\frac65$ , $\frac53$ , and $\frac56$ , in that order; the square root is power $\frac12$ . Enter those powers as the list $a$ , then type $100((1+p/100)^a-1)$ . Desmos shows approximately $[700,14603.34,102300,3100]$ . Only the last result matches the $c=3100$ you found, so $c=100\left(\left(1+\frac{p}{100}\right)^{5/6}-1\right)$ . Choice D.

Approach 2: Prove the interval rule with exponents
Step 1
Find the multiplier for 6 input units
No Desmos needed. Exponent rules give the exact relationship for every allowed $b$ . A multiplier is the new output divided by the old output. Write that ratio for an increase of $6$ in $x$ :
$\frac{f(x+6)}{f(x)}=\frac{240b^{(x+6)/2}}{240b^{x/2}}.$
Cancel the matching $240$ factors:
$\frac{f(x+6)}{f(x)}=\frac{b^{(x+6)/2}}{b^{x/2}}.$
Subtract exponents when dividing powers of the same base:
$\frac{f(x+6)}{f(x)}=b^{(x+6)/2-x/2}.$
Combine the halves:
$\frac{f(x+6)}{f(x)}=b^3.$
Step 2
Translate the given percent into a multiplier
An increase of $p\%$ keeps the whole original amount, represented by $1$ , and adds $p/100$ more. So the 6-unit multiplier you found must satisfy
$b^3=1+\frac{p}{100}.$
Step 3
Find the multiplier for 5 input units
Increasing $x$ by $5$ raises the exponent $x/2$ by $5/2$ . Subtract the exponents in the ratio of new output to old output:
$\frac{f(x+5)}{f(x)}=b^{(x+5)/2-x/2}.$
Combine the halves:
$\frac{f(x+5)}{f(x)}=b^{5/2}.$
The $5/2$ is an exponent change, not yet a percent increase.
Step 4
Rewrite the 5-unit factor using the 6-unit factor
The power rule multiplies exponents when you raise a power to another power. Since $3\cdot\frac56=\frac52$ , rewrite the 5-unit factor in terms of $b^3$ :
$b^{5/2}=(b^3)^{5/6}.$
Replace $b^3$ with the given 6-unit multiplier:
$b^{5/2}=\left(1+\frac{p}{100}\right)^{5/6}.$
The fraction is $5/6$ because the desired input interval is $5$ units out of the original $6$ .
Step 5
Turn the factor into the requested percent
An increase of $c\%$ has multiplier $1+c/100$ , so set it equal to the 5-unit multiplier:
$1+\frac{c}{100}=\left(1+\frac{p}{100}\right)^{5/6}.$
Subtract $1$ to keep only the increase:
$\frac{c}{100}=\left(1+\frac{p}{100}\right)^{5/6}-1.$
Multiply by $100$ to express that increase as a percent:
$c=100\left(\left(1+\frac{p}{100}\right)^{5/6}-1\right).$
That's the percent increase for every 5-unit rise in $x$ . Choice D.

Question 165·200 Super-Hard SAT Math Questions·Advanced Math

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