Solve for x log base 2 of x+ log base 2 of x-6=4

Problem

(log_2)(x)+(log_2)(x−6)=4

Solution

1. Apply the product rule for logarithms, which states that (log_b)(M)+(log_b)(N)=(log_b)(M*N)

(log_2)(x*(x−6))=4

2. Rewrite in exponential form to remove the logarithm, using the definition (log_b)(y)=a⇔ba=y

x*(x−6)=2

3. Simplify and expand the equation to form a standard quadratic equation.

x2−6*x=16

4. Rearrange the equation into the form a*x2+b*x+c=0 by subtracting 16 from both sides.

x2−6*x−16=0

5. Factor the quadratic expression by finding two numbers that multiply to −16 and add to −6

(x−8)*(x+2)=0

6. Solve for x by setting each factor equal to zero.

x=8

x=−2

7. Check for extraneous solutions by ensuring the arguments of the original logarithms are positive. Since (log_2)(−2) and (log_2)(−2−6) are undefined, we discard x=−2

x=8

Final Answer

(log_2)(x)+(log_2)(x−6)=4⇒x=8

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