Question #08d03fe4 - SAT Suite Question Bank

ID: 08d03fe4

Modded SAT Question Bank by Abdullah Mallik

For the exponential function $f$ , the value of $f left parenthesis 1 right parenthesis$ is $k$ , where $k$ is a constant. Which of the following equivalent forms of the function $f$ shows the value of $k$ as the coefficient or the base?

$f (x) = 50 {(2)}^{x + 1}$
$f left parenthesis x right parenthesis equals 80 left parenthesis 2 right parenthesis Superscript x$
$f (x) = 128 {(2)}^{x - 1}$
$f (x) = 205 {(2)}^{x - 2}$

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Correct Answer: C

Rationale

Choice C is correct. For the form of the function in choice C, $f (x) = 128 {(1.6)}^{x - 1}$ , the value of $f left parenthesis 1 right parenthesis$ can be found as $128 left parenthesis 1.6 right parenthesis Superscript 1 minus 1$ , which is equivalent to $128 left parenthesis 1.6 right parenthesis Superscript 0$ , or $128$ . Therefore, $k equals 128$ , which is shown in $f (x) = 128 {(1.6)}^{x - 1}$ as the coefficient.

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.

Question Difficulty: Hard