AI Math Solver
Resources
Questions
Pricing
Login
Register
Home
>
Questions
>
Solve Logarithmic Equation: 3 log(3x) - log x = log(9x+4) + log3
Mathematics
Grade 10 (Junior High)
Question Content
Solve the equation: 3 log(3x) - log x = log(9x + 4) + log 3
Correct Answer
x = 4
Detailed Solution Steps
1
Step 1: Apply the power rule to 3 log(3x): log((3x)^3) = log(27x^3).
2
Step 2: Apply the quotient rule to the left side: log(27x^3 / x) = log(27x^2).
3
Step 3: Apply the product rule to the right side: log[(9x + 4)*3] = log(27x + 12).
4
Step 4: Set the arguments equal (one-to-one property): 27x^2 = 27x + 12.
5
Step 5: Rearrange into standard quadratic form: 27x^2 - 27x - 12 = 0 → 9x^2 - 9x - 4 = 0.
6
Step 6: Factor the quadratic: (3x - 4)(3x + 1) = 0. Solutions are x=4/3 or x=-1/3.
7
Step 7: Check for valid solutions (logarithm arguments must be positive). x=-1/3 makes 3x negative, so discard. x=4/3 is valid. Wait correction: 27x^2=27x+12 → 27x²-27x-12=0 divide by 3:9x²-9x-4=0, quadratic formula x=(9±√(81+144))/18=(9±√225)/18=(9±15)/18. x=(24)/18=4/3, x=(-6)/18=-1/3. Correct, x=4/3 is valid.
Knowledge Points Involved
1
Power Rule of Logarithms
k log_a m = log_a(m^k) for real k, positive m, a>0, a≠1. Moves coefficients as exponents inside logarithms.
2
Product and Quotient Rules of Logarithms
Product rule: log_a m + log_a n = log_a(mn); Quotient rule: log_a m - log_a n = log_a(m/n). Combine or split logarithms with the same base.
3
Validity of Logarithm Arguments
For log_a m to be defined, m > 0, a > 0, a ≠ 1. Always check solutions to ensure all logarithm arguments are positive.
Loading solution...
