Question

what is the approximate value of q in the equation below?
$q + \log_{2}6 = 2q + 2$

• $-1.613$
• $-1.522$
• $0.585$
• $3.079$

Explanation:

Step1: Isolate q terms

Subtract \( q \) and \( 2 \) from both sides: \( \log_{2}6 - 2 = q \)

Step2: Calculate \( \log_{2}6 \)

Using change of base formula: \( \log_{2}6=\frac{\ln 6}{\ln 2}\approx\frac{1.7918}{0.6931}\approx2.585 \)

Step3: Compute q

Substitute \( \log_{2}6\approx2.585 \): \( q = 2.585 - 2 = 0.585 \)? Wait, no—wait, original equation: \( q + \log_{2}6 = 2q + 2 \). Let's re - arrange correctly:
Subtract \( q \) and \( 2 \) from both sides: \( \log_{2}6 - 2 = q \)? Wait, no: \( q + \log_{2}6=2q + 2\)
Subtract \( q \) from both sides: \( \log_{2}6=q + 2\)
Subtract \( 2 \) from both sides: \( q=\log_{2}6 - 2\)
Calculate \( \log_{2}6=\frac{\ln 6}{\ln 2}\approx\frac{1.791759}{0.693147}\approx2.58496\)
Then \( q = 2.58496-2 = 0.58496\approx0.585\)

Wait, but let's check again. Wait, maybe I made a mistake in sign. Wait, the equation is \( q+\log_{2}6 = 2q + 2\)
Let's move all q terms to one side: \( \log_{2}6-2=2q - q\)
So \( q=\log_{2}6 - 2\)
\( \log_{2}6=\log_{2}(2\times3)=\log_{2}2+\log_{2}3 = 1+\log_{2}3\approx1 + 1.58496=2.58496\)
Then \( q=2.58496 - 2=0.58496\approx0.585\)

Answer:

0.585 (corresponding to the option "0.585")