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# 一些log的題目 <急>

設log10<2>= a log10<3>= b 以a.b表示下列各式

1. log10<18>

2.log10<根號6>

3.log3<4>

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求x

1.log10(x-2)+log10<5>=2

2.log3(x+2)+log3(2x-1)=1

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算下列各式

1.log2<3>*log3<16>

2.log4<3>*log9<5>*log125<16>

3.(log2<3>+log4<9>)(log3<2>+log9<8>)

### 1 Answer

- TonyLv 51 decade agoFavorite Answer
<第一部分>

log2 = a, log3 = b (以10為底的常用對數可以直接寫成log，把底數省略)

1. log18 = log(2*9) = log(2*3^2) = log2 + log(3^2) = log2 + 2log3 = a+2b

2. log(根號6) = log(6^(1/2)) = (1/2)*log6 = (1/2)*log(2*3) = (1/2)*(log2+log3) = (1/2)*(a+b)

3. log3<4> (這題考的是換底公式 loga<b> = (logc<a>)/(logc<b>))

log3<4> = log4/log3 = (log(2^2))/log3 = (2log2)/log3 = 2a/b

<第二部分>

1. log(x-2)+log5 = 2

log(x-2)*5 = log(10^2)

=> (兩邊同時拿掉log) 5(x-2) = 100

=> x-2 = 20

=> x = 22

2. log3(x+2)+log3(2x-1) = 1

log3((x+2)(2x-1)) = log3<3>

=> (兩邊同時拿掉log) 2x^2+3x-2 = 3

=> 2x^2+3x-5 = 0

=> (2x+5)(x-1) = 0

x = -5/2(不和，真數必須大於0) or x = 1 (這個是我們要的答案)

<第三部分>

這一部分考的是連鎖率，它其實是換底公式移項後的結果

loga<b>*logb<c> = loga<c>

這個式子兩邊同除以loga<b>後會得到換底公式

1. log2<3>*log3<16> = log2<16> = log2(2^4) = 4log2<2> = 4

2. log4<3>*log9<5>*log125<16> = (log2^2<3>)*(log3^2<5>)*(log5^3<16>)

當底數有次方數時，將它提出來要變成倒數(loga^k<b> = (1/k)*loga<b>)

所以:

原式 = ((1/2)log2<3>)*((1/2)log3<5>)*((1/3)log5<16>)

=(1/12)log2<3>*log3<5>*log5<16> (接著套用連鎖率)

=(1/12)log2<16> = (1/12)log2(2^4) = 4/12 = 1/3

3. (log2<3>+log4<9>)(log3<2>+log9<8>)

=(log2<3>+log2^2(3^2))(log3<2>+log3^2(2^3))

=(log2<3>+(2/2)log2<3>)(log3<2>+(3/2)log3<2>)

=2log2<3>*(5/2)log3<2> = 5*log2<3>*log3<2> = 5log2<2> = 5