指數與對數計算詳解

(1)2^127－1是一個質數，此質數是幾位數?
Sol
a=2^127－1
log(2^127－1)～log2^127=127log2=38.2257
log2=0.3010 =>2是1位數
log20=1.3010=1+0.3010 =>20是2位數
log200=2.3010=2+0.3010 =>200是3位數
So
2^127－1是39位數

(2)(1/5)^50在小數點第幾位才出現不為0的數字?
Sol
a=(1/5)^50
loga=50log(1/5)=50(log2-log10)=50*(0.3010-1)=－34.95=－35+0.05
log0.011=－2+log1.1 =>在小數點第2位才出現不為0的數字
log0.0011=－3+log1.1 =>在小數點第3位才出現不為0的數字
So
(1/5)^50在小數點第35位才出現不為0的數字

(3)設a=log5_2，b=log2_3，則log18=?
Sol
a=log2/log5=log2/(1-log2)
a-alog2=log2
a=(1+a)log2
log2=a/(1+a)
b=log3/log2
log3=blog2=ab(1+a)
log18=log2+2log3
=a/(1+a)+2ab(1+a)
=(a+2ab)/(a+1)

(4)logx首數與log345的首數相同，logx的尾數與log12.5的尾數相同，則x=?
Sol
log345=log100+log3.45=2+log3.45
log12.5=log10+log1.25=1+log1.25
logx=2+log1.25=log125
x=125

(5)已知a為整數，且2.2>loga>2.1，則a^3為幾位數?
Sol
2.2>loga>2.1
10^2.2>a>10^2.1
10^6.6>a^3<10^6.3
6+1=7
a^3為7位數?

(6)設x為實數，且6^x－3^x+2^x－1=0，則6^x=?
Sol
6^x－3^x+2^x－1=0
(6^x－3^x)+(2^x－1)=0
3^x(2^x－1)+(2^x－1)=0
(3^x+1)(2^x－1)=0
2^x－1=0
2^x=1
x=0
6^x=6^0=1

(7)若log3_x+log3_y=2，則1/x+1/y之最小值為何?
Sol
x>0,y>0
log3_(xy)=2
xy=9
(1/x+1/y)/2>=√[(1/x)*(1/y)]
(1/x+1/y)>=2√[(1/x)*(1/y)]=2√(1/9)
(1/x+1/y)>=2/3

(8)設n為自然數，且x=3^n+3^n+3^n，則下列何者等於x^2?
Sol
x=3*3^n=3^(n+1)
x^2=3^(2n+2)

(9)若log2_(x^2－x－1)之值恆為正，則x之範圍為何?
Sol
log2_(x^2－x－1)>0
x^2－x－1>1
x^2－x－2>0
(x－2)(x+1)>0
x－2>0 or x+1<0
x>2 orx<－1

(11)設x、y為正實數，若2log(x－2y)=logx+logy，則x/y之值為何?
Sol
2log(x－2y)=logx+logy
(x－2y)^2=xy
x^2－4xy+4y^2=xy
x^2－5xy+4y^2=0
(x－y)(x－4y)=0
(1) x=y
X－2y=－x<0(不合)
(2) x=4y
x/y\=4

11.
x-2y>0
==>x>2y
2log(x-2y)=logx+logy
==>log(x-2y)^2=logxy
==>(x-2y)^2=xy
==>(x-4y)(x-y)=0
==>x=4y或(x=y不合因x>2y)

x/y=4y/y=4

2015-01-11 14:18:43 補充：
(3)設a=log5(底數)2，b=log2(底數)3，則log18=? 答：a+2ab/a+1
a=log5(底數)2=log2/log5
==>alog5=log2
==>a(1-log2)=log2
==>log2=a/(a+1)

b=log2(底數)3=log3/log2
==>blog2=log3

log18=log3^2+log2=2log3+log2=2blog2+log2=(2b+1)log2=a(2b+1)/(a+1)
=(a+2ab)/(a+1)

2015-01-11 14:19:14 補充：
(7)若log3(底數)x+log3(底數)y=2，則1/x+1/y之最小值為何?? 答：2/3
log3(底數)x+log3(底數)y=2
==>log3_xy=log3_3^2
==>xy=9
1/x+1/y>=2((1/x)*(1/y))^(1/2)=(1/9)^(1/2) 算幾不等式
==>1/x+1/y>=2/3
其最小值為2/3

2015-01-11 14:39:51 補充：
(6)設x為實數，且6^X-3^X+2^X-1=0，則6^X=? 答：1
令2^X=u,3^X=v
6^X-3^X+2^X-1=0
==>(3^X)(2^X)-3^X+2^X-1=0
==>uv-v+u-1=0
==>u(v+1)-(v+1)=0
==>(u-1)(v+1)=0
==>u=1,(v=-1不合因真數要大於0)
2^X=1 ==>X=0

6^X=6^0=1

2015-01-11 14:43:47 補充：
(8)設n為自然數，且x=3^n+3^n+3^n，則下列何者等於x^2? 答：3^(2n+2)
x=3^n+3^n+3^n
==>x=3*(3^n)=3^(n+1)
==>x^2=(3^(n+1))^2=3^(2n+2)