Solve for x log of x+ log of x-48=2

Problem

(log_)(x)+(log_)(x−48)=2

Solution

1. Apply the product rule for logarithms, which states that (log_)(a)+(log_)(b)=(log_)(a*b)

(log_)(x*(x−48))=2

2. Rewrite in exponential form by using the base of the common logarithm, which is 10

x*(x−48)=10

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

x2−48*x=100

x2−48*x−100=0

4. Factor the quadratic equation by finding two numbers that multiply to −100 and add to −48

(x−50)*(x+2)=0

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

x=50

x=−2

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

x=50

Final Answer

(log_)(x)+(log_)(x−48)=2⇒x=50

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