高中數學 對數(log)問題part2【每題皆要有詳解！】

Question

1.解下列方程式

(1)logx+log2=2

(2)log(x+3)-log(x-1)=1

(3)log(10^x+100)=x/2+1+log2

2.解不等式

(1)-1<log_1/3 x<0

(2)log_3 (x^2+2x)>1

(3)log_1/2 (x-1)+log_1/2 (x-3)≥-3

3.已知log2≒0.3010、log3≒0.4771，若正整數n滿足logn≒1.23，則n=？

4.將下列各數表示成科學記號。b×10^n，其中1≤b<10，n為整數

(1)3^50

(2)(2/3)^20

(3)7^30

Answer 1

最佳解答

1.解下列方程式(1)logx+log2=2log(x)=2*log(10)-log(2)=log(10^2)-log(2)=log(100/2)=log(50)x=50

(2)log(x+3)-log(x-1)=1log[(x+3)/(x-1)]=log(10)(x+3)=10(x-1)0=-x-3+10x-10x=13/9

(3)log(10^x+100)=x/2+1+log2=log(10)^[(x+2)/2]+log2=log{10)^[(x+2)/2]2}10^x+100=10^[(x+2)/2]2=2*[10^(x/2)*10]=20*10^(x/2)0=y^2-20y+100.....y=10^(x/2)=(y-10)^210=y=10^(x/2)x=2

2.解不等式(1)-1<log_(1/3)x<0-1<log(x)/log(1/3)<0-1<log(x)/[-log(3)]<01>log(x)/log(3)>0log(3)>log(x)>log(1)3>x>1

(2)log_3 (x^2+2x)>1log(x^2+2x)/log(3)>1log(x^2+2x)>log(3)x^2+2x-3>0(x+3)(x-1)>0x<-3 or x>1

(3)log_1/2 (x-1)+log_1/2 (x-3)≥-3log(x-1)/log(1/2)+log(x-3)/log(1/2)>=-3log[(x-1)(x-3)]/log(1/2)>=-3log[(x-1)(x-3)]/[-log(2)]>=-3log[(x-1)(x-3)]/log(2)<=3log[(x-1)(x-3)]<=3*log(2)(x-1)(x-3)<=8x^2-4x+3-8<=0x^2-4x-5<=0(x-5)(x+1)<=0-1<=x<=5

3.已知log2≒0.3010、log3≒0.4771，若正整數n滿足log(n)≒1.23，則n=？log(n)=0.4771*2+0.3010=1.2552≒1.23=2*log(3)+log(2)=log(9*2)=log(18)n=18 => log(18)非常接近1.23取n=17 => log(17)=1.23更接近1.23ans: n=17

4.將下列各數表示成科學記號。b×10^n，其中1≤b<10，n為整數(1) x=3^50log(x)=50*log(3)=50*0.30103=15.0514997832=15*log10+0.0514997832=15*log10+log(1.1259).....使用小算盤計算=log(1.1259*10^15)x=1.1259*10^15

(2) y=(2/3)^20log(y)=20*(log(2)-log(3))=20*(0.30103-0.47712)=-3.5218251811=-3.5218251811+4-4=-4+0.4781748189=-4*log(10)+log(3.0073).....使用小算盤計算=log(3.0073/10^4)y=3.0073*10^(-4)

(3) z=7^30log(z)=30*log(7)=30*0.84509804=25.3529412004=25*log(10)+log(2.254).....使用小算盤計算z=2.254*10^25