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Ln(x)+ln(x-2)=ln(x+2)+ln(x-3)
According to the law of logarithm;
Ln(a) + Ln(b) = Ln(ab)
Sp,
Ln[(x)*(x -2)] = Ln[(x + 2)*(x - 3)
Taking anti-log natural on both sides
Anti-Ln *Ln[(x)*(x -2)] = anti-Ln *Ln[(x + 2)*(x - 3)
x* (x - 2) = (x + 2) * (x - 3)
x2 - 2x = x2 + 2x -3x - 6
x2 - 2x = x2 - x - 6
x2 - 2 x - x2 + x + 6 = 0
- x + 6 = 0
x = 6
x = 6
The value of x is 6.
According to the law of logarithm;
Ln(a) + Ln(b) = Ln(ab)
Sp,
Ln[(x)*(x -2)] = Ln[(x + 2)*(x - 3)
Taking anti-log natural on both sides
Anti-Ln *Ln[(x)*(x -2)] = anti-Ln *Ln[(x + 2)*(x - 3)
x* (x - 2) = (x + 2) * (x - 3)
x2 - 2x = x2 + 2x -3x - 6
x2 - 2x = x2 - x - 6
x2 - 2 x - x2 + x + 6 = 0
- x + 6 = 0
x = 6
x = 6
The value of x is 6.
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