中四數學(對數方程) 15點

(a)log(2x) + 1 = log (5x – 10)log(2x) + log10 = log (5x –10)log [(2x)(10)] = log (5x –10)(2x)(10) = 5x - 1020x = 5x - 1020x – 5x = -1015x = -103x = -2x = -2/3
(No real solution since you can’t takea log of negative number) Check:log(2(-2/3) + 1 = log [5(-2/3)– 10)]log(-4/3) + log10 = log [-10/3– 10)]log[(-4/3)10] = log [(-10/3) –10][(-4/3)10] = [-10/3 – 10)]-40/3 =[-10/3 – 30/3)]-40/3 =(-10-30)/3 -40/3 = - 40/3It is OK (b)log2(x+5) – 3 = log2(4 – x)log2(x+5) – 3log2 (2) = log2(4 – x)log2(x+5) – log2 [(2)^3] = log2(4 – x)log2(x+5) – log2 [8] = log2(4 – x)log2 [(x+5)/8] = log2(4 – x)(x + 5)/8 = 4 – xx + 5 = 8(4 – x)x + 5 = 32 – 8xx + 8x = 32 – 59x = 27x = 27/9 = 3x = 3 (answer)

Check:log2(3+5) – 3 = log2(4 – 3)log2(8) – 3 = log2(1)log2(2^3) – 3 = 03log2(2) – 3 = 03(1) – 3 = 00 = 0 It is OK.

It is because you have answered b) wrongly.

1. log (2x) + 1 = log (5x – 10) 2x + log10 = log (5x – 10) 2x (10) = 5x – 10 20x = 5x – 10 15x = -10 x = -2/3 (No real solution) 2. log 2 (x – 5) – 3 = log 2 (4 – x) (x – 5) = 4 – x + 3 x – 5 = -x + 7 2x = 12 x = 6

2011-02-12 11:39:50 補充：
What is the aim of 002 to submit the identical answers plus adding just only the checking??