Solve for x log base 6 of x+ log base 6 of x-5=2

Problem

(log_6)(x)+(log_6)(x−5)=2

Solution

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

(log_6)(x*(x−5))=2

2. Rewrite in exponential form by using the definition of a logarithm (log_b)(y)=a⇔ba=y

x*(x−5)=6

3. Expand and simplify the equation to form a standard quadratic equation.

x2−5*x=36

x2−5*x−36=0

4. Factor the quadratic equation by finding two numbers that multiply to −36 and add to −5

(x−9)*(x+4)=0

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

x=9

x=−4

6. Check for extraneous solutions by ensuring the arguments of the original logarithms are positive. Since (log_6)(−4) is undefined, we discard x=−4

x=9

Final Answer

(log_6)(x)+(log_6)(x−5)=2⇒x=9

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