標題:
發問:
1. Let s= log 2 and t= log 3, express each of the following in terms of s and t. (a) log25 (b) log1.44^(2/3) 2. Solve the equation 4^x= 20 - 4^(x-1). 3. Solve the equation 3^(x+2) - 3(x+1) = 18. 4. log(5x-9) / log (x-1) = 2. 5. Solve the equation log(3^x + 3^-x) + log 3 = 1.
最佳解答:
1. Let s= log 2 and t= log 3．express each of the following in terms of s and t. (a) log25 =log(100/4) =log100－log4 =2－2log2 =2－2s (b) log1.44^(2/3) =(2/3)log(1.44) =(2/3)log(144/100) =(2/3)[log144－log100] =(2/3)[log(2^6)+log(3^2)－] =(2/3)(6s+2t－2) =(12s+4t－4)/3 2. Solve the equation 4^x= 20－4^(x－1). Sol 4*4^x=80－4^x 5*4^x=80 4^x=16 x=2 3. Solve the equation 3^(x+2)－3(x+1) =18. Sol 9*3^x－*3^x=18 6*3^x=18 3^x=3 x=1 4. log(5x－9) /log (x－1) =2. log(5x－9)=2log(x－1) 5x－9=(x－1)^2 x^2－2x+1=5x－9 x^2－7x+10=0 (x－2)(x－5)=0 x=2(不合) or x=5 5. Solve the equation log(3^x + 3^－x) + log3 = 1 Sol Set 3^x=y log(3^x+3^(－x)).+log3=1 3*(3^x+3^(－x))=10 3*(y+1/y)=10 3y^2+3=10y 3y^2－10y+3=0 (3y－1)(y－3)=0 y=1/3 or y=3 (1) y=1/3 =>x=－1 (2) y=3 =>x=1
其他解答:
1.(a) s= log 2 and t= log 3 log25 =log(100/4) =log100-log4 =2-2log2 =2-2s (b) log1.44^(2/3) =(2/3)log(144/100) =(2/3)log(144)-(2/3)log(100) =(2/3)log(12)^2-(4/3) =(4/3)log(4*3)-(4/3) =(4/3)[log4+log3-1] =(4/3)(2log2+log3-1) =(4/3)(2s+t-1) 2. 4^x= 20 - 4^(x-1) 4^x+ 4^(x-1)= 20 4^x+ 4^(x)(4)= 20 4^x (5)= 20 4^x=4 x=1 3.the equation shoud be corrected into the following 3^(x+2) - 3^(x+1) = 18 3^(x)(9) - 3^(x)(3) = 18 3^(x)(6)=18 3^(x)=3 x=1 4.log(5x-9) / log (x-1) = 2. log(5x-9)= 2log (x-1) log(5x-9) = log (x-1)^2 log(5x-9) -log (x-1)^2=0 log[(5x-9)/(x-1)^2]=0 [(5x-9)/(x-1)^2}=1 (5x-9)=(x-1)^2 5x-9=x^2-2x+1 x^2-7x+8=0 (x+1)(x-8)=0 x+1=0 or x-8=0 x=-1(rejected) or x=8 x=8 2009-11-08 12:04:47 補充： log(5x-9) / log (x-1) = 2. log(5x-9)= 2log (x-1) log(5x-9) = log (x-1)^2 log(5x-9) -log (x-1)^2=0 log[(5x-9)/(x-1)^2]=0 [(5x-9)/(x-1)^2}=1 (5x-9)=(x-1)^2 5x-9=x^2-2x+1 x^2-7x+10=0 (x-2)(x-5)=0 x-2=0 or x-5=0 x=2(rejected) or x=5 x=5
此文章來自奇摩知識+如有不便請留言告知
F.4 log questions(20分)發問:
1. Let s= log 2 and t= log 3, express each of the following in terms of s and t. (a) log25 (b) log1.44^(2/3) 2. Solve the equation 4^x= 20 - 4^(x-1). 3. Solve the equation 3^(x+2) - 3(x+1) = 18. 4. log(5x-9) / log (x-1) = 2. 5. Solve the equation log(3^x + 3^-x) + log 3 = 1.
最佳解答:
1. Let s= log 2 and t= log 3．express each of the following in terms of s and t. (a) log25 =log(100/4) =log100－log4 =2－2log2 =2－2s (b) log1.44^(2/3) =(2/3)log(1.44) =(2/3)log(144/100) =(2/3)[log144－log100] =(2/3)[log(2^6)+log(3^2)－] =(2/3)(6s+2t－2) =(12s+4t－4)/3 2. Solve the equation 4^x= 20－4^(x－1). Sol 4*4^x=80－4^x 5*4^x=80 4^x=16 x=2 3. Solve the equation 3^(x+2)－3(x+1) =18. Sol 9*3^x－*3^x=18 6*3^x=18 3^x=3 x=1 4. log(5x－9) /log (x－1) =2. log(5x－9)=2log(x－1) 5x－9=(x－1)^2 x^2－2x+1=5x－9 x^2－7x+10=0 (x－2)(x－5)=0 x=2(不合) or x=5 5. Solve the equation log(3^x + 3^－x) + log3 = 1 Sol Set 3^x=y log(3^x+3^(－x)).+log3=1 3*(3^x+3^(－x))=10 3*(y+1/y)=10 3y^2+3=10y 3y^2－10y+3=0 (3y－1)(y－3)=0 y=1/3 or y=3 (1) y=1/3 =>x=－1 (2) y=3 =>x=1
其他解答:
1.(a) s= log 2 and t= log 3 log25 =log(100/4) =log100-log4 =2-2log2 =2-2s (b) log1.44^(2/3) =(2/3)log(144/100) =(2/3)log(144)-(2/3)log(100) =(2/3)log(12)^2-(4/3) =(4/3)log(4*3)-(4/3) =(4/3)[log4+log3-1] =(4/3)(2log2+log3-1) =(4/3)(2s+t-1) 2. 4^x= 20 - 4^(x-1) 4^x+ 4^(x-1)= 20 4^x+ 4^(x)(4)= 20 4^x (5)= 20 4^x=4 x=1 3.the equation shoud be corrected into the following 3^(x+2) - 3^(x+1) = 18 3^(x)(9) - 3^(x)(3) = 18 3^(x)(6)=18 3^(x)=3 x=1 4.log(5x-9) / log (x-1) = 2. log(5x-9)= 2log (x-1) log(5x-9) = log (x-1)^2 log(5x-9) -log (x-1)^2=0 log[(5x-9)/(x-1)^2]=0 [(5x-9)/(x-1)^2}=1 (5x-9)=(x-1)^2 5x-9=x^2-2x+1 x^2-7x+8=0 (x+1)(x-8)=0 x+1=0 or x-8=0 x=-1(rejected) or x=8 x=8 2009-11-08 12:04:47 補充： log(5x-9) / log (x-1) = 2. log(5x-9)= 2log (x-1) log(5x-9) = log (x-1)^2 log(5x-9) -log (x-1)^2=0 log[(5x-9)/(x-1)^2]=0 [(5x-9)/(x-1)^2}=1 (5x-9)=(x-1)^2 5x-9=x^2-2x+1 x^2-7x+10=0 (x-2)(x-5)=0 x-2=0 or x-5=0 x=2(rejected) or x=5 x=5
文章標籤
全站熱搜
留言列表
