Solve for x

Problem

(log_5)(x−7)+(log_5)(x−4)−(log_5)(x)=1

Solution

1. Apply the product rule for logarithms to the first two terms, which states (log_b)(M)+(log_b)(N)=(log_b)(M*N)

(log_5)((x−7)*(x−4))−(log_5)(x)=1

2. Apply the quotient rule for logarithms, which states (log_b)(M)−(log_b)(N)=(log_b)(M/N)

(log_5)(((x−7)*(x−4))/x)=1

3. Convert to exponential form using the definition (log_b)(y)=a⇔ba=y

((x−7)*(x−4))/x=5

4. Multiply by x to clear the fraction and expand the binomials.

(x−7)*(x−4)=5*x

x2−11*x+28=5*x

5. Rearrange into a quadratic equation by subtracting 5*x from both sides.

x2−16*x+28=0

6. Factor the quadratic by finding two numbers that multiply to 28 and add to −16

(x−14)*(x−2)=0

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

x=14

x=2

8. Check for extraneous solutions by ensuring the arguments of the original logarithms are positive. For x=2 the term (log_5)(2−7) is undefined. For x=14 all arguments are positive.

x=14

Final Answer

(log_5)(x−7)+(log_5)(x−4)−(log_5)(x)=1⇒x=14

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