Question

use the table to estimate one solution, to the nearest tenth, of this equation:

$-3x + 5 = 5^x$

the values in the table have been rounded to the nearest tenth.

$x$	$-3x + 5$	$5^x$
$0$	$5$	$1$
$1$	$2$	$5$
$0.6$	$3.2$	$2.6$
$0.7$	$2.9$	$3.1$
$0.65$	$3.1$	$2.8$

$x \approx \square$

Explanation:

Step1: Understand the equation

We need to find \( x \) such that \( -3x + 5 = 5^x \). This means we look for \( x \) where the values of \( -3x + 5 \) and \( 5^x \) are equal (or very close, since values are rounded).

Step2: Analyze the table values

• For \( x = 0.6 \): \( -3x + 5 = 3.2 \), \( 5^x = 2.6 \) ( \( 3.2>2.6 \) )
• For \( x = 0.7 \): \( -3x + 5 = 2.9 \), \( 5^x = 3.1 \) ( \( 2.9<3.1 \) )
• For \( x = 0.65 \): \( -3x + 5 = 3.1 \), \( 5^x = 2.8 \) ( \( 3.1>2.8 \) )

We see that between \( x = 0.6 \) and \( x = 0.7 \), the value of \( -3x + 5 \) decreases from \( 3.2 \) to \( 2.9 \) and \( 5^x \) increases from \( 2.6 \) to \( 3.1 \). At \( x = 0.7 \), \( 5^x = 3.1 \) and \( -3x + 5 = 2.9 \), the difference is smaller. Also, at \( x = 0.7 \), \( 5^x = 3.1 \) and \( -3x + 5 = 2.9 \), which is closer than at \( x = 0.6 \). Since we need to the nearest tenth, and at \( x = 0.7 \), the two functions are closer (difference of \( 0.2 \)) compared to \( x = 0.6 \) (difference of \( 0.6 \)). Also, when \( x = 0.7 \), \( 5^x = 3.1 \) and \( -3x + 5 = 2.9 \), and as \( x \) increases, \( 5^x \) grows and \( -3x + 5 \) decreases. The crossing happens around \( x = 0.7 \) (since at \( x = 0.7 \), \( 5^x \) is slightly above \( -3x + 5 \), and at \( x = 0.65 \), \( -3x + 5 \) is above \( 5^x \)). But looking at the rounded values, at \( x = 0.7 \), \( 5^x = 3.1 \) and \( -3x + 5 = 2.9 \), and the closest tenth where they are near equal (considering rounding) is \( x = 0.7 \) (since the change from \( x = 0.6 \) to \( 0.7 \) shows that \( 5^x \) overtakes \( -3x + 5 \) around \( 0.7 \), and the values are closest at \( x = 0.7 \) when rounded to the nearest tenth).

Answer:

\( x \approx 0.7 \)