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Solve Logarithmic Equation: log2 + 3logx + log(1/x²) = 2log(2x)
Mathematics
Grade 10 (Junior High)
Question Content
Solve the equation: log 2 + 3 log x + log(1/x²) = 2 log(2x)
Correct Answer
x = 8
Detailed Solution Steps
1
Step 1: Apply power rule to each term: log2 + log(x³) + log(x^(-2)) = log((2x)^2).
2
Step 2: Simplify log(x^(-2)) = -2logx, and log((2x)^2)=log(4x²).
3
Step 3: Apply product rule to left side: log(2 * x³ * x^(-2)) = log(2x).
4
Step 4: Set arguments equal: 2x = 4x².
5
Step 5: Rearrange: 4x² -2x=0 → 2x(2x-1)=0. Solutions x=0 or x=1/2.
6
Step 6: Check validity: x=0 makes logx undefined, so discard. x=1/2 is valid (all arguments positive). Wait correction: log2 +3logx +log(1/x²)=log2 + logx³ + logx^(-2)=log(2*x³*x^(-2))=log(2x). Right side 2log(2x)=log((2x)^2)=log(4x²). So log(2x)=log(4x²) →2x=4x² →4x²-2x=0→2x(2x-1)=0→x=0 or x=1/2. Correct, x=1/2 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. Applies to positive and negative coefficients, converting them to exponents.
2
Product Rule of Logarithms
log_a m + log_a n = log_a(mn) for positive m,n, a>0, a≠1. Combines multiple added logarithms into a single logarithm of the product of all arguments.
3
Negative Exponents in Logarithms
log_a(m^(-k)) = -k log_a m, which is equivalent to log_a(1/m^k). Used to simplify logarithms of reciprocal terms.
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