Question

the solution to the equation \\(\log_{2}x + \log_{2}(x - 6) = 4\\) is \\(x = \square\\)

Explanation:

Step1: Apply log addition rule

Using the property of logarithms \(\log_a M+\log_a N = \log_a(MN)\), we can rewrite the left - hand side of the equation \(\log_2x+\log_2(x - 6)=4\) as \(\log_2[x(x - 6)]=4\).

Step2: Convert logarithmic to exponential form

Recall that if \(\log_a b=c\), then \(b = a^c\). For our equation \(\log_2[x(x - 6)] = 4\), we have \(x(x - 6)=2^4\).

Step3: Solve the quadratic equation

Expand the left - hand side: \(x^2-6x = 16\).
Rearrange it to the standard quadratic form \(x^2-6x - 16=0\).
Factor the quadratic equation: We need two numbers that multiply to \(- 16\) and add up to \(-6\). The numbers are \(-8\) and \(2\). So, \(x^2-6x - 16=(x - 8)(x + 2)=0\).
Set each factor equal to zero: \(x-8 = 0\) or \(x + 2=0\). This gives us two solutions \(x = 8\) or \(x=-2\).

Step4: Check the domain of the logarithmic function

For the logarithm \(\log_2x\) and \(\log_2(x - 6)\) to be well - defined, we need \(x>0\) and \(x - 6>0\) (i.e., \(x>6\)).
When \(x=-2\), \(\log_2(-2)\) and \(\log_2(-2 - 6)=\log_2(-8)\) are not defined (since the argument of a logarithm must be positive).
When \(x = 8\), \(\log_28=3\) and \(\log_2(8 - 6)=\log_22 = 1\), and \(3 + 1=4\), which satisfies the original equation.

Answer:

\(8\)