Solve for x log of x+3+ log of x+4 = log of 20

Problem

(log_)(x+3)+(log_)(x+4)=(log_)(20)

Solution

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

(log_)((x+3)*(x+4))=(log_)(20)

2. Remove the logarithms by using the property that if (log_)(u)=(log_)(v) then u=v

(x+3)*(x+4)=20

3. Expand the product on the left side of the equation using the FOIL method.

x2+7*x+12=20

4. Rearrange the equation into standard quadratic form a*x2+b*x+c=0 by subtracting 20 from both sides.

x2+7*x−8=0

5. Factor the quadratic expression by finding two numbers that multiply to −8 and add to 7

(x+8)*(x−1)=0

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

x=−8

x=1

7. Check for extraneous solutions by substituting the values back into the original logarithmic expressions. Since the argument of a logarithm must be positive, x=−8 is invalid because (log_)(−8+3)=(log_)(−5) is undefined.

x=1⇒(log_)(1+3)+(log_)(1+4)=(log_)(4)+(log_)(5)=(log_)(20)

Final Answer

(log_)(x+3)+(log_)(x+4)=(log_)(20)⇒x=1

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