Question

question 5
solve for ( x ):
( log_{2}(x + 6)=2log_{2}(x) )
( circ ) -2
( circ ) 3
( circ ) 3.2
( circ ) 2
( circ ) 3, -2

Explanation:

Step1: Use logarithm power rule

The power rule of logarithms states that \( n\log_b(a)=\log_b(a^n) \). Apply this to the right - hand side of the equation \( \log_2(x + 6)=2\log_2(x) \). We get \( \log_2(x + 6)=\log_2(x^2) \).

Step2: Use the one - to - one property of logarithms

If \( \log_b(m)=\log_b(n) \), then \( m = n \) (for \( b>0,b
eq1,m>0,n>0 \)). So from \( \log_2(x + 6)=\log_2(x^2) \), we can conclude that \( x + 6=x^2 \).

Step3: Rearrange the equation to a quadratic form

Rearrange \( x + 6=x^2 \) to get a quadratic equation \( x^2-x - 6=0 \).

Step4: Factor the quadratic equation

Factor \( x^2 - x-6 \). We need two numbers that multiply to - 6 and add up to - 1. The numbers are - 3 and 2. So \( x^2 - x - 6=(x - 3)(x+2)=0 \).

Step5: Solve for x

Set each factor equal to zero:

• If \( x - 3=0 \), then \( x = 3 \).
• If \( x+2=0 \), then \( x=-2 \).

Step6: Check the domain of the logarithm function

The argument of a logarithm \( \log_b(u) \) must satisfy \( u>0 \).

• For \( x=-2 \): The argument of \( \log_2(x) \) is \( x=-2 \), which is less than 0. So \( x = - 2 \) is not in the domain of the original logarithmic equation.
• For \( x = 3 \): The argument of \( \log_2(x+6) \) is \( 3 + 6=9>0 \) and the argument of \( \log_2(x) \) is \( 3>0 \). So \( x = 3 \) is in the domain.

Answer:

B. 3