Solve for x log of x+ log of x-21=2

Problem

(log_)(x)+(log_)(x−21)=2

Solution

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

(log_)(x*(x−21))=2

2. Rewrite in exponential form by using the definition of a common logarithm (base 10).

x*(x−21)=10

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

x2−21*x=100

x2−21*x−100=0

4. Factor the quadratic by finding two numbers that multiply to −100 and add to −21

(x−25)*(x+4)=0

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

x=25

x=−4

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

x=25

Final Answer

(log_)(x)+(log_)(x−21)=2⇒x=25

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