Question

what is the approximate value of q in the equation?
$q + \log_{2} 6 = 2q + 2$
3.079
0.585
−1.613
−1.522

Explanation:

Step1: Rearrange the equation

We start with the equation \( q + \log_{2}6 = 2q + 2 \). To solve for \( q \), we can subtract \( q \) and \( 2 \) from both sides.

$$ q + \log_{2}6 - q - 2 = 2q + 2 - q - 2 $$

Simplifying both sides, we get:

$$ \log_{2}6 - 2 = q $$

Step2: Calculate the value of \( \log_{2}6 \)

We know that \( \log_{2}6=\frac{\ln 6}{\ln 2} \) (by the change - of - base formula \( \log_{a}b = \frac{\ln b}{\ln a} \)).
We calculate \( \ln 6\approx1.7918 \) and \( \ln 2\approx0.6931 \).
So \( \log_{2}6=\frac{1.7918}{0.6931}\approx2.585 \).

Step3: Calculate the value of \( q \)

Now we substitute the value of \( \log_{2}6 \) into the equation for \( q \):
\( q=\log_{2}6 - 2\approx2.585-2 = 0.585 \)? Wait, no, wait. Wait, let's recalculate. Wait, \( \log_{2}6=\frac{\log_{10}6}{\log_{10}2}=\frac{0.7782}{0.3010}\approx2.585 \). Then \( q=\log_{2}6 - 2\approx2.585 - 2=0.585 \)? Wait, no, wait the original equation is \( q+\log_{2}6 = 2q + 2 \), so moving terms: \( \log_{2}6-2=2q - q\), so \( q=\log_{2}6 - 2 \). Wait, \( \log_{2}6\approx2.58496 \), then \( 2.58496-2 = 0.58496\approx0.585 \). Wait, but let's check again. Wait, maybe I made a mistake in the sign. Wait, the equation is \( q+\log_{2}6=2q + 2 \). Subtract \( q \) from both sides: \( \log_{2}6=q + 2 \). Then subtract 2 from both sides: \( q=\log_{2}6 - 2 \). Yes, that's correct. So \( \log_{2}6\approx2.585 \), so \( q = 2.585-2=0.585 \).

Answer:

\( 0.585 \) (the option with 0.585)