Question

exponential and logarithmic functions
table for an exponential function
the function g is defined by the following rule.
$g(x) = 3^x$
find $g(x)$ for each $x$-value in the table.

$x$	$g(x)$
$0$	$square$
$1$	$square$
$2$	$square$
$3$	$square$

Explanation:

Step1: Calculate for \( x = -1 \)

Substitute \( x = -1 \) into \( g(x) = 3^x \). Using the exponent rule \( a^{-n} = \frac{1}{a^n} \), we get \( g(-1) = 3^{-1} = \frac{1}{3} \).

Step2: Calculate for \( x = 0 \)

Substitute \( x = 0 \) into \( g(x) = 3^x \). Using the rule \( a^0 = 1 \) (for \( a
eq 0 \)), we get \( g(0) = 3^0 = 1 \).

Step3: Calculate for \( x = 1 \)

Substitute \( x = 1 \) into \( g(x) = 3^x \). We get \( g(1) = 3^1 = 3 \).

Step4: Calculate for \( x = 2 \)

Substitute \( x = 2 \) into \( g(x) = 3^x \). We get \( g(2) = 3^2 = 9 \).

Step5: Calculate for \( x = 3 \)

Substitute \( x = 3 \) into \( g(x) = 3^x \). We get \( g(3) = 3^3 = 27 \).

Answer:

For \( x = -1 \), \( g(x) = \frac{1}{3} \); for \( x = 0 \), \( g(x) = 1 \); for \( x = 1 \), \( g(x) = 3 \); for \( x = 2 \), \( g(x) = 9 \); for \( x = 3 \), \( g(x) = 27 \)

Filling the table:

\( x \)	\( g(x) \)
\( 0 \)	\( 1 \)
\( 1 \)	\( 3 \)
\( 2 \)	\( 9 \)
\( 3 \)	\( 27 \)