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標題:
F.4 MATHS (logarithmic)
發問:
Find the unknown 1) 3^2x+1+3^2x-1=8(cor. to 3 s.f.) 2) 圖片參考：http://imgcld.yimg.com/8/n/HA00292709/o/701011160144213873411420.jpg 3) 3 log9 2 - log9 54 –log9 36 Simplify the equation 1) log3 (3^(x+1) -18)=x plz show me the steps ( better if including explanations )thz
1) 32??1 + 32??1 = 8 32(32??1) + 32??1 = 8 32??1(32 + 1) = 8 32??1(10) = 8 log[32??1(10)] = log8 log32??1 + log10 = log8 (2x - 1)log3 = -1 + log8 2x - 1 = (-1 + log8)/log3 2x = 1 + [(-1 + log8)/log3] x = (1/2) + [(-1 + log8)/2log3] x = 0.398 ===== 2) logy = log(2x) - 1 …… (1) logx = 2log√(y + 1) …… (2) (1): logy = log2x - log10 logy = log(2x/10) y = 2x/10 x = 5y ….. (3) (2): logx = 2log(y + 1)1?2 logx = (2)(1/2)log(y + 1) logx = log(y + 1) x = y + 1 …… (4) (3) = (4): 5y = y + 1 4y = 1 y = 0.25 Put y = 0.25 into (4): x = 0.25 + 1 x = 1.25 Hence, x = 1.25, y = 0.25 ===== 3) 3log92 - log954 - log936 = log923 - log954 - log936 = log98 - log954 - log936 = log8/log9 - log54/log9 - log36/log9 = (log8 - log54 - log36)/log9 = log[8/(54*36)]/log9 = log(1/243)/log9 = -log243/log9 = -log(3)?/log(3)2 = -5log3/2log3 = -5/2 = -2.5 ==== 1) log3(3??1 - 18) = x 3??1 - 18 = 3? 3??1 - 3? = 18 (3)3? - 3? = 18 3?(3 - 1) = 18 3?(2) = 18 3? = 9 3? = 32 x = 2
其他解答:
2. From the first equation, logy = log2x - 1 logy = log2x - log10 logy = log(x/5) y = x/5---------------------(1) From the second equation, logx = 2logroot(y + 1) logx = log(y + 1) x = y + 1------------------(2) Soving (1) and (2), x = 5/4, y = 1/4|||||F.4 MATHS (logarithmic)Find the unknown 1) 3^2x+1+3^2x-1=8(cor. to 3 s.f.) 2) 圖片參考：http://imgcld.yimg.com/8/n/HA00749348/o/701011160144213873411420.jpg 3) 3 log9 2 - log9 54 –log9 36 Simplify the equation 1) log3 (3^(x+1) -18)=x Ans: 1) 3^2x+1+3^2x-1=8 (2x+1)log3+(2x-1)log3=log8 4xlog3=log8 x=log8/4log3 =0.473 (cor. to. 3 s.f.) 2) logy=log(2x)-1 log(y/2x)=-1 y/2x=0.1 y=0.2x .....(1) 2logx=2log(y+1)^0.5 log x^2=log(y+1) x^2=y+1 y=(x^2)-1 ....(2) (1)=(2):y=0.2x=(x^2)-1 (x^2)-0.2x-1=0 5(x^2)-x-5=0 ....(3) x= -0.905(rej.) or 1.10 (cor .to 3 s.f.) Put x =1.10 into (1), y=0.220 (cor. to 3 s. f.) (x,y)=(1.10,0.220) 3)3 log9 2 - log9 54 –log9 36 =log9 8-log9 54-log 936 =log9(8/(54*36)) =log9 (1/243) =-5log3/2log3 =-2.5 log3 (3^(x+1) -18)=x 3^(x+1) -18=3^x 3^(x+1) -3^x-18=0 2(3^x)=18 3^x=9 x=2|||||2. y=log(2x)-1+log log x=2log log(2x)-1+log logx =3log(2x)-1+log logx=6log x=6
F.4 MATHS (logarithmic)
發問:
Find the unknown 1) 3^2x+1+3^2x-1=8(cor. to 3 s.f.) 2) 圖片參考：http://imgcld.yimg.com/8/n/HA00292709/o/701011160144213873411420.jpg 3) 3 log9 2 - log9 54 –log9 36 Simplify the equation 1) log3 (3^(x+1) -18)=x plz show me the steps ( better if including explanations )thz
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最佳解答:1) 32??1 + 32??1 = 8 32(32??1) + 32??1 = 8 32??1(32 + 1) = 8 32??1(10) = 8 log[32??1(10)] = log8 log32??1 + log10 = log8 (2x - 1)log3 = -1 + log8 2x - 1 = (-1 + log8)/log3 2x = 1 + [(-1 + log8)/log3] x = (1/2) + [(-1 + log8)/2log3] x = 0.398 ===== 2) logy = log(2x) - 1 …… (1) logx = 2log√(y + 1) …… (2) (1): logy = log2x - log10 logy = log(2x/10) y = 2x/10 x = 5y ….. (3) (2): logx = 2log(y + 1)1?2 logx = (2)(1/2)log(y + 1) logx = log(y + 1) x = y + 1 …… (4) (3) = (4): 5y = y + 1 4y = 1 y = 0.25 Put y = 0.25 into (4): x = 0.25 + 1 x = 1.25 Hence, x = 1.25, y = 0.25 ===== 3) 3log92 - log954 - log936 = log923 - log954 - log936 = log98 - log954 - log936 = log8/log9 - log54/log9 - log36/log9 = (log8 - log54 - log36)/log9 = log[8/(54*36)]/log9 = log(1/243)/log9 = -log243/log9 = -log(3)?/log(3)2 = -5log3/2log3 = -5/2 = -2.5 ==== 1) log3(3??1 - 18) = x 3??1 - 18 = 3? 3??1 - 3? = 18 (3)3? - 3? = 18 3?(3 - 1) = 18 3?(2) = 18 3? = 9 3? = 32 x = 2
其他解答:
2. From the first equation, logy = log2x - 1 logy = log2x - log10 logy = log(x/5) y = x/5---------------------(1) From the second equation, logx = 2logroot(y + 1) logx = log(y + 1) x = y + 1------------------(2) Soving (1) and (2), x = 5/4, y = 1/4|||||F.4 MATHS (logarithmic)Find the unknown 1) 3^2x+1+3^2x-1=8(cor. to 3 s.f.) 2) 圖片參考：http://imgcld.yimg.com/8/n/HA00749348/o/701011160144213873411420.jpg 3) 3 log9 2 - log9 54 –log9 36 Simplify the equation 1) log3 (3^(x+1) -18)=x Ans: 1) 3^2x+1+3^2x-1=8 (2x+1)log3+(2x-1)log3=log8 4xlog3=log8 x=log8/4log3 =0.473 (cor. to. 3 s.f.) 2) logy=log(2x)-1 log(y/2x)=-1 y/2x=0.1 y=0.2x .....(1) 2logx=2log(y+1)^0.5 log x^2=log(y+1) x^2=y+1 y=(x^2)-1 ....(2) (1)=(2):y=0.2x=(x^2)-1 (x^2)-0.2x-1=0 5(x^2)-x-5=0 ....(3) x= -0.905(rej.) or 1.10 (cor .to 3 s.f.) Put x =1.10 into (1), y=0.220 (cor. to 3 s. f.) (x,y)=(1.10,0.220) 3)3 log9 2 - log9 54 –log9 36 =log9 8-log9 54-log 936 =log9(8/(54*36)) =log9 (1/243) =-5log3/2log3 =-2.5 log3 (3^(x+1) -18)=x 3^(x+1) -18=3^x 3^(x+1) -3^x-18=0 2(3^x)=18 3^x=9 x=2|||||2. y=log(2x)-1+log log x=2log log(2x)-1+log logx =3log(2x)-1+log logx=6log x=6
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