maths問題............log! 超難!唔識!HELP

1. 10^(2logy)
= 10^(log y^2)
= y^2

2. log 10^y
= (log10)(y)
= y

3. log(16y ^2)-2log(40y)
= log (16y^2 / 1600y^2)
= log (1/100)
= -2

4. log base a(b)
= log b / log a
= 1 over (log a / log b)
= 1 over log base b(a)

5. log base 3(X) = log base 9(3X - 2)
log (X) / log 3 = log (3X - 2) / 2log 3
2log (X) = log (3X - 2)
log (X^2) = log (3X - 2)
X^2 - 3X + 2 = 0
X = 1 , 2

2007-04-14 21:34:28 補充：
因為你本書垃圾= =買過本low中4書有嫁~~

1)Let x=10^(2logy)
logx=log[10^2logy]
logx=2logy*log10
logx=log(y^2)
x=y^2
2) log10^ y
=y*(log10)
=y
3) log(16y ^2)-2log(40y)
=log(16y ^2)-[log(40y)^2]
=log(16y^2/1600y^2)
=log(1/100)= -2
4)Below prove that log basic a(b)= log basic c(b) / log basic c(a) where c is any constant:
letc^x=b,c^y=a,then c=a^(1/y),put this into c^x=b --> a^[(1/y)*x]=b,so thata^(x/y)=b
notice that x=log basic c(b),y=log basic c(a) ,x/y= log basic a(b)
that is log basic a(b)= log basic c(b) / log basic c(a)
Back to ur Q, log basic a(b)
=logb/loga=1/(loga/logb)=1/log basic b(a) here take c=10
5)By using wt we get in 4),-----log basic a(b)= log basic c(b) / log basic c(a)
LHS=logx/log3,RHS=log(3x-2)/log9
logx/log3=log(3x-2)/log3^2
logx/log3=log(3x-2)/2log3
2logx=log(3x-2)
logx^2=log(3x-2)
x^2-3x+2=0
x=1 or 2

1) 10^2logy
= log10^2logy
= 2logy x log10
= 2logy x 1
(log basic 10(10) = 1)
= 2logy

2) log10^y = ylog10

3) log(16y ^2)-2log(40y)
= 2log(16y)-2log(40y)
= 2log(16y over 40y)
= 2log(2 over 5)
= -0.79588.....
= -0.796 (3s.f.)

4) log basic a(b)
= log basic b(b) over log basic b(a)
(change of base rule)
= 1 over log basic b(a)
(log basic b(b) = 1 as b^1=b)

5) log basic 3(X) = log basic 9(3X - 2)
log basic 3(X) = log basic 3(3X - 2)^(1 over 2)
(3^2=9 so 9^(1 over 2)=3)
X = (3X - 2)^(1 over 2)
X^2 = 3X - 2
X^2 - 3X + 2 = 0
(X - 2)(X - 1) = 0
X= 1 or 2

我諗應該係咁做，希望幫到你(又冇答錯)la！ ^^

2007-04-14 21:18:04 補充：
2) log10^y = ylog10(log basic 10(10) = 1)=y~打漏左~

1. 10^(2logy)
= 10^(log y^2)
= y^2

2. log 10^y
= y

3. log(16y ^2)-2log(40y)
= log (16y^2 / 1600y^2)
= log (1/100)
= -2

4. log base a(b)
= log b / log a
= 1 over (log a / log b)
= 1 over log base b(a)

5. log base 3(X) = log base 9(3X - 2)
log (X) / log 3 = log (3X - 2) / 2log 3
2log (X) = log (3X - 2)
log (X^2) = log (3X - 2)
X^2 - 3X + 2 = 0
X = 1 , 2