Question

which equation can you solve to find the potential solutions to the equation \\(\log_{2}x + \log_{2}(x - 6) = 4\\)?

Explanation:

Step1: Recall log product rule

The logarithmic product rule states that $\log_b M + \log_b N=\log_b(MN)$. Apply this to $\log_2 x+\log_2(x - 6)$.
$$\log_2 x+\log_2(x - 6)=\log_2[x(x - 6)]$$

Step2: Set equal to 4

The original equation is $\log_2 x+\log_2(x - 6)=4$, so substitute the left - hand side with the result from Step 1:
$$\log_2[x(x - 6)] = 4$$

Step3: Convert to exponential form

Using the definition of a logarithm: if $\log_b a=c$, then $a = b^c$. Here, $b = 2$, $a=x(x - 6)$ and $c = 4$. So we get:
$$x(x - 6)=2^4$$
Simplify $2^4=16$, so the equation becomes $x(x - 6)=16$, which expands to $x^{2}-6x - 16 = 0$. But the question is about which equation we can solve to find the potential solutions. After applying the log product rule and converting to exponential form, the equation we solve is $x(x - 6)=16$ (or $x^{2}-6x-16 = 0$). But if we follow the steps of combining the logs first:

First, combine the logarithms: $\log_2(x(x - 6))=4$. Then convert to exponential form: $x(x - 6)=2^4=16$. So the equation we solve to find the potential solutions is $x(x - 6)=16$ (or expanding, $x^{2}-6x - 16=0$). But the key step is using the log property to get $\log_2[x(x - 6)]=4$ and then converting to $x(x - 6)=16$.

Answer:

The equation we can solve is $x(x - 6)=16$ (or $x^{2}-6x - 16 = 0$). If we consider the step of combining the logs and converting to exponential form, the equation derived is $x(x - 6)=16$ (equivalent to $x^{2}-6x-16 = 0$) which we solve to find the potential solutions of the original logarithmic equation.