標題:
Maths
發問:
3^2x-3^x-6=0 3(3^2x0-10(3^x)=3=0 (logx)^2-logx-12=0 2x-7√x+6=0 x-4√x=-3 i^18=? i^43=?
Hi ! I am lop****** , feel happy to answer your question. Q 1 : 3^2x-3^x-6=0 A 1 : 3^2x-3^x-6=0 Substitute u = e^(x^2) into the left hand side: u^2-u-6 = 0 Add 6 to both sides: u^2-u = 6 Add 1/4 to both sides: u^2-u+1/4 = 25/4 Factor the left hand side: (u-1/2)^2 = 25/4 Take the square root of both sides: |(u-1/2)| = 5/2 Eliminate the absolute value: u-1/2 = -5/2 or u-1/2 = 5/2 Add 1/2 to both sides: u = -2 or u-1/2 = 5/2 Substitute back for u = 3^x: 3^x = -2 or u-1/2 = 5/2 Take the logarithm to the base 3 of both sides: x = (iπ+log(2))/(log(3)) or u-1/2 = 5/2 Add 1/2 to both sides: x = (iπ+log(2))/(log(3)) or u = 3 Substitute back for u = 3^x: x = (iπ+log(2))/(log(3)) or 3^x = 3 Take the logarithm to the base 3 of both sides: x = (iπ+log(2))/(log(3)) or x = 1 Answer : x = 1 or (iπ+log(2))/(log(3)) Q 2 : 3(3^2x0-10(3^x)=3=0 A 2 : What do you mean of ' =3=0 ' and ' 3^2x0 ' ? Q 3 : (logx)^2-logx-12=0 A 3 : (logx)^2-logx-12=0 Substitute u = log(x): u^2-u-12 = 0 Factor the left hand side: (u+3)(u-4)=0 u = -3 or 4 Substitute back for u = log(x): log(x) = -3 or log(x) = 4 Cancel logarithms by taking exp of both sides: x = 1/e^3 or x = e^4 Now test that these solutions are appropriate by substitution into the original equation: Check the solution x = 1/e^3: log^2(x)-log(x)-12 => -12-log(1/e^3)+log^2(1/e^3) = 0 So the solution is correct. Check the solution x = e^4: log^2(x)-log(x)-12 => -12-log(e^4)+log^2(e^4) = 0 So the solution is correct. Answer : x = 1/e^3 or e^4 . Q 4 : 2x-7√x+6=0 A 4 : 2x-7√x+6=0 Subtract 2x+6 from both sides: -7√x = -2x-6 Divide both sides by -7: √x = 1/7 (2x+6) Square both sides: x = [(2x+6)^2]/49 Expand out terms on the right hand side: x = (4 x^2)/49+(24 x)/49+36/49 Subtract ((4 x^2)/49+(24 x)/49+36/49) from both sides: -(4x^2)/49+(25 x)/49-36/49 = 0 49[-(4x^2)/49+(25 x)/49-36/49] = 0 -4x^2 +25x-36 = 0 -(-4x^2+25x-36)=0 4x^2-25x+36=0 (4x-9)(x-4)=0 4x=9 or x=4 x= 9/4 or 4 Answer : x = 9/4 or 4 ====================================== 由於字數限制，Q5 , Q6 及 Q7 將到意見欄回答，不便之處，敬請原諒。 2011-07-21 16:38:45 補充： Q 5 : x-4√x=-3 A 5 : x - 4√x = -3 Subtract x from both sides: -4√x = -x-3 Divide both sides by -4: √x = (x+3)/4 Square both sides: x = [(x+3)^2]/16 Expand out terms on the right hand side: x = x^2/16+(3x)/8+9/16 2011-07-21 16:38:57 補充： Subtract (x^2/16+(3x)/8+9/16) from both sides: -x^2/16+(5x)/8-9/16=0 16[-x^2/16+(5x)/8-9/16]=0 -x^2+10x-9=0 -(-x^2+10x-9)=0 x^2-10x+9=0 (x-1)(x-9)=0 x=1or9 Answer : x = 1 or 9 . ====================================== 2011-07-21 16:39:05 補充： Q 6 : i^18=? A 6 : i^18 = i^16 × i^2 = (i^4)^4 × ( i × i ) = (i×i×i×i)^4 × (-1) = (-1×-1)^4 × (-1) = 1^4 × (-1) = 1 × (-1) = -1 Answer : i^18 = -1 . ====================================== 2011-07-21 16:39:31 補充： Q 7 : i^43=? A 7 : i^43 = i^40 × i^3 = (i^4)^10 × (i×i×i) = (i×i×i×i)^10 × (-1×i) = - [ (-1×-1)^10 × 1 × i ] = - ( 1^10 × i ) = - ( 1 × i ) = -i Answer : i^43 = -i . ======================================
其他解答:
Put u = 3^x 3^2x - 3^x - 6 = 0 (3^x)^2 - (3^x) - 6 = 0 u^2 - u - 6 = 0 (u - 3)(u + 2) = 0 u = 3 or u = -2 3^x = 3 or 3^x = -2 (rejected) 3^x = 3^1 x = 1 ===== Put u = 3^x 3(3^2x) - 10(3^x) + 3 = 0 3(3^x)^2- 10(3^x) + 3 = 0 3u^2- 10u + 3 = 0 (u - 3)(3u - 1) = 0 u = 3 or u = 1/3 3^x = 3 or 3^x = 1/3 3^x = 3^1 or 3^x = 3^-1 x = 1 or x = -1 ===== Put u = logx (logx)^2 - logx - 12 = 0 u^2- u - 12 = 0 (u - 4)(u + 3) = 0 u = 4 or u = -3 logx = 4 or logx = -3 x = 10^4 or x = 10^-3 x = 10000 or x = 1/1000 ===== Put u = √x 2x - 7√x + 6 = 0 2u^2- 7u + 6 = 0 (2u - 3)(u - 2) = 0 u = 3/2 or u = 2 √x = 3/2 or √x = 2 x = 9/4 or x = 4 ===== Put u = √x x - 4√x = -3 u2 - 4u + 3 = 0 (u - 1)(u - 3) = 0 u = 1 or u = 3 √x = 1 or √x = 3 x = 1 or x = 9 ===== i = √-1 i^2 = (√-1)^2 = -1 i^3 = i^2*i = -i i^4 = (i^2)^2 = (-1)^2 = 1 i^18 = (i^16)*(i^2) = (i^4)^4 * (i^2) = (1)^4 * (-1) = -1 ==== i =√-1 i^2 = (√-1)^2 = -1 i^3 = i^2*i = -i i^4 = (i^2)^2 = (-1)^2 = 1 i^43 = (i^40) * (i^3) = (i^4)^10 * (i^3) = (1)^10 * (-i) = -i
Maths
發問:
3^2x-3^x-6=0 3(3^2x0-10(3^x)=3=0 (logx)^2-logx-12=0 2x-7√x+6=0 x-4√x=-3 i^18=? i^43=?
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最佳解答:Hi ! I am lop****** , feel happy to answer your question. Q 1 : 3^2x-3^x-6=0 A 1 : 3^2x-3^x-6=0 Substitute u = e^(x^2) into the left hand side: u^2-u-6 = 0 Add 6 to both sides: u^2-u = 6 Add 1/4 to both sides: u^2-u+1/4 = 25/4 Factor the left hand side: (u-1/2)^2 = 25/4 Take the square root of both sides: |(u-1/2)| = 5/2 Eliminate the absolute value: u-1/2 = -5/2 or u-1/2 = 5/2 Add 1/2 to both sides: u = -2 or u-1/2 = 5/2 Substitute back for u = 3^x: 3^x = -2 or u-1/2 = 5/2 Take the logarithm to the base 3 of both sides: x = (iπ+log(2))/(log(3)) or u-1/2 = 5/2 Add 1/2 to both sides: x = (iπ+log(2))/(log(3)) or u = 3 Substitute back for u = 3^x: x = (iπ+log(2))/(log(3)) or 3^x = 3 Take the logarithm to the base 3 of both sides: x = (iπ+log(2))/(log(3)) or x = 1 Answer : x = 1 or (iπ+log(2))/(log(3)) Q 2 : 3(3^2x0-10(3^x)=3=0 A 2 : What do you mean of ' =3=0 ' and ' 3^2x0 ' ? Q 3 : (logx)^2-logx-12=0 A 3 : (logx)^2-logx-12=0 Substitute u = log(x): u^2-u-12 = 0 Factor the left hand side: (u+3)(u-4)=0 u = -3 or 4 Substitute back for u = log(x): log(x) = -3 or log(x) = 4 Cancel logarithms by taking exp of both sides: x = 1/e^3 or x = e^4 Now test that these solutions are appropriate by substitution into the original equation: Check the solution x = 1/e^3: log^2(x)-log(x)-12 => -12-log(1/e^3)+log^2(1/e^3) = 0 So the solution is correct. Check the solution x = e^4: log^2(x)-log(x)-12 => -12-log(e^4)+log^2(e^4) = 0 So the solution is correct. Answer : x = 1/e^3 or e^4 . Q 4 : 2x-7√x+6=0 A 4 : 2x-7√x+6=0 Subtract 2x+6 from both sides: -7√x = -2x-6 Divide both sides by -7: √x = 1/7 (2x+6) Square both sides: x = [(2x+6)^2]/49 Expand out terms on the right hand side: x = (4 x^2)/49+(24 x)/49+36/49 Subtract ((4 x^2)/49+(24 x)/49+36/49) from both sides: -(4x^2)/49+(25 x)/49-36/49 = 0 49[-(4x^2)/49+(25 x)/49-36/49] = 0 -4x^2 +25x-36 = 0 -(-4x^2+25x-36)=0 4x^2-25x+36=0 (4x-9)(x-4)=0 4x=9 or x=4 x= 9/4 or 4 Answer : x = 9/4 or 4 ====================================== 由於字數限制，Q5 , Q6 及 Q7 將到意見欄回答，不便之處，敬請原諒。 2011-07-21 16:38:45 補充： Q 5 : x-4√x=-3 A 5 : x - 4√x = -3 Subtract x from both sides: -4√x = -x-3 Divide both sides by -4: √x = (x+3)/4 Square both sides: x = [(x+3)^2]/16 Expand out terms on the right hand side: x = x^2/16+(3x)/8+9/16 2011-07-21 16:38:57 補充： Subtract (x^2/16+(3x)/8+9/16) from both sides: -x^2/16+(5x)/8-9/16=0 16[-x^2/16+(5x)/8-9/16]=0 -x^2+10x-9=0 -(-x^2+10x-9)=0 x^2-10x+9=0 (x-1)(x-9)=0 x=1or9 Answer : x = 1 or 9 . ====================================== 2011-07-21 16:39:05 補充： Q 6 : i^18=? A 6 : i^18 = i^16 × i^2 = (i^4)^4 × ( i × i ) = (i×i×i×i)^4 × (-1) = (-1×-1)^4 × (-1) = 1^4 × (-1) = 1 × (-1) = -1 Answer : i^18 = -1 . ====================================== 2011-07-21 16:39:31 補充： Q 7 : i^43=? A 7 : i^43 = i^40 × i^3 = (i^4)^10 × (i×i×i) = (i×i×i×i)^10 × (-1×i) = - [ (-1×-1)^10 × 1 × i ] = - ( 1^10 × i ) = - ( 1 × i ) = -i Answer : i^43 = -i . ======================================
其他解答:
Put u = 3^x 3^2x - 3^x - 6 = 0 (3^x)^2 - (3^x) - 6 = 0 u^2 - u - 6 = 0 (u - 3)(u + 2) = 0 u = 3 or u = -2 3^x = 3 or 3^x = -2 (rejected) 3^x = 3^1 x = 1 ===== Put u = 3^x 3(3^2x) - 10(3^x) + 3 = 0 3(3^x)^2- 10(3^x) + 3 = 0 3u^2- 10u + 3 = 0 (u - 3)(3u - 1) = 0 u = 3 or u = 1/3 3^x = 3 or 3^x = 1/3 3^x = 3^1 or 3^x = 3^-1 x = 1 or x = -1 ===== Put u = logx (logx)^2 - logx - 12 = 0 u^2- u - 12 = 0 (u - 4)(u + 3) = 0 u = 4 or u = -3 logx = 4 or logx = -3 x = 10^4 or x = 10^-3 x = 10000 or x = 1/1000 ===== Put u = √x 2x - 7√x + 6 = 0 2u^2- 7u + 6 = 0 (2u - 3)(u - 2) = 0 u = 3/2 or u = 2 √x = 3/2 or √x = 2 x = 9/4 or x = 4 ===== Put u = √x x - 4√x = -3 u2 - 4u + 3 = 0 (u - 1)(u - 3) = 0 u = 1 or u = 3 √x = 1 or √x = 3 x = 1 or x = 9 ===== i = √-1 i^2 = (√-1)^2 = -1 i^3 = i^2*i = -i i^4 = (i^2)^2 = (-1)^2 = 1 i^18 = (i^16)*(i^2) = (i^4)^4 * (i^2) = (1)^4 * (-1) = -1 ==== i =√-1 i^2 = (√-1)^2 = -1 i^3 = i^2*i = -i i^4 = (i^2)^2 = (-1)^2 = 1 i^43 = (i^40) * (i^3) = (i^4)^10 * (i^3) = (1)^10 * (-i) = -i
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