Solve for x log of 3x+7+ log of x-2=1

Problem

(log_)(3*x+7)+(log_)(x−2)=1

Solution

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

(log_)((3*x+7)*(x−2))=1

2. Rewrite in exponential form by assuming the base is 10 (common logarithm).

(3*x+7)*(x−2)=10

3. Expand the product on the left side of the equation.

3*x2−6*x+7*x−14=10

4. Combine like terms and set the quadratic equation to zero by subtracting 10 from both sides.

3*x2+x−24=0

5. Factor the quadratic equation to find the potential values for x

(3*x−8)*(x+3)=0

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

3*x−8=0⇒x=8/3

x+3=0⇒x=−3

7. Check for extraneous solutions by ensuring the arguments of the original logarithms are positive.

If *x=−3,(log_)(−3−2)=(log_)(−5)* (undefined)

If *x=8/3,(log_)(3*(8/3)+7)=(log_)(15)* and *(log_)(8/3−2)=(log_)(2/3)* (valid)

Final Answer

x=8/3

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