Solve for x log base 8 of x+ log base 8 of x-12=2

Problem

(log_8)(x)+(log_8)(x−12)=2

Solution

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

(log_8)(x*(x−12))=2

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

x*(x−12)=8

3. Simplify the equation by distributing x and calculating the square of 8.

x2−12*x=64

4. Set the quadratic equation to zero by subtracting 64 from both sides.

x2−12*x−64=0

5. Factor the quadratic expression by finding two numbers that multiply to −64 and add to −12

(x−16)*(x+4)=0

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

x=16

x=−4

7. Check for extraneous solutions by ensuring the arguments of the original logarithms are positive. Since (log_8)(−4) and (log_8)(−4−12) are undefined, we discard x=−4

x=16

Final Answer

(log_8)(x)+(log_8)(x−12)=2⇒x=16

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