1. 鳳山高級中學 B1 3-3 數學科 試卷___年 ___班 座號：___ 姓名：__________
一、單選題
（ ）1.設 a﹐ 為正實數﹐
b 已知 log 7 a = 11 ﹐log 7 b = 13 ﹐ log 7 (a + b) 的值最接近下列哪個選項？ (1)12 (2)13
則
(3)14 (4)23 (5)24﹒
解答 2
解析 log 7 a = 11 ﹐得 a = 711 ﹐ log 7 b = 13 ﹐得 b = 713 ﹐又 a + b = 711 + 713 = 711 (1 + 49) = 711 ⋅ 50 ≈ 713 ﹐得
log 7 (a + b) ≈ log 7 713 = 13 ﹐故選(2)﹒
（ ）2.在坐標平面上﹐ P為 y = −( x + 1)( x − 2) 圖形上的一點﹐ P的x坐標為log310﹐
設 若 試問P的位置在哪一象限？
(1)一 (2)二 (3)三 (4)四 (5)不一定﹒
解答 4
解析 因點 P 的 x 坐標為 log 3 10 > 2 ﹐且 y 坐標為 y = −(log 3 10 + 1)(log 3 10 − 2) < 0 ﹐
知點 P 在第四象限﹐故選(4)﹒
二、多選題
（ ）3.若a﹐b > 0﹐a≠1﹐則下列敘述何者正確？ (1) log0.60.7 = log67 (2) log 6
7 = log67 (3) log1821 = log67
6
(4) 7log3 6 = 6log3 7 (5)若log64a = log128b﹐則logab = ﹒
7
解答 24
三、計算題
4.求下列各式之值：
49 1
(1) log 6 + log 6 ﹒
4 441
25 7 1 1
(2) log10 − log10 − log10 + log10 49 ﹒
9 4 9 2
16 1 5 2
(3) log10 + log10 − log10 − 3log10 2 ﹒
5 2 6 3
1
解答 (1) −2 ;(2)2;(3) −
2
49 1 72 1 1
解析 (1)原式 = log 6 ( × ) = log 6 ( × 2 2 ) = log 6 ( ) 2 = −2 ﹒
4 441 4 7 ×3 6
25 4
(2)原式 = log10 ( × × 9 × 49) = log10 100 = 2 ﹒
9 7
16 5 3 1 1 1
(3)原式 = log10 ( × × × ) = log10 =− ﹒
5 6 2 8 10 2
－ 1 －

2. 5.求(log2)3 + (log5)3 + (log2)(log125)之值﹒
解答 1
解析 原式= (log2)3 + (log5)3 + log2．log125 = (log2)3 + (log5)3 + 3．log2．log5
= (log2)3 + (log5)3 + 3．log2．log5．(log2 + log5) = (log2 + log5)3 = (log10)3 = 1
6.設 log a α = log b β = log ab
10 ﹐已知 α ≠ β ﹐則 αβ =____________﹒
解答 100
k
解析 令 log a α = log b β = log ab
10 = k ﹐ α = a k ﹐ β = b k ﹐ 10 = ( ab ) k = ab 2 ﹐
k
αβ = a k ⋅ b k = ( ab) k = [( ab) 2 ]2 = 102 = 100 7.小康想使用三個 2 與數學符號來表示一實數﹐試問 log 2 log 2 2 所表示
的實數﹒
解答 −3
1 1 1 1 1
1
解析 因 2 = [(2 2 ) 2 ] 2 = 2 8 ﹐所求實數為 log 2 log 2 2 8 = log 2 = −3 ﹒
8
8.設 4logx − 3．xlog2 − 4 = 0﹐則x = ____________﹒
解答 100
解析 (2log x)2 − 3．2log x − 4 = 0 ⇒ (2log x − 4)(2log x + 1) = 0 ⇒ 2log x = 4 = 22 ⇒ log x = 2 ∴ x = 100
9.設 18a = 2﹐試以a表示log32 = ____________﹒
2a
解答
1− a
log 3
解析 18a = 2 ⇒ a log18 = log 2 ⇒ a (2log 3 + log 2) = log 2 ⇒ 2a ． + a =1
log 2
1− a 2a
⇒ log 2 3 = ∴ log 3 2 =
2a 1− a
1 1 1 1
10.設m﹐n為正整數﹐log2(1 + ) + log2(1 + ) + log2(1 + ) + … + log2(1 + ) = log2n﹐則 2m − 3n = -
m m +1 m+2 m + n −1
____________﹒
解答 −2
1 1 1 1
解析 log2(1 + ) + log2(1 + ) + log2(1 + ) + … + log2(1 + )
m m +1 m+2 m + n −1
m +1 m+2 m+3 m+n
= log2( ) + log2( ) + log2( ) + … + log2( )
m m +1 m+2 m + n −1
m +1 m + 2 m + 3 m+n m+n
= log2( × × × … × ) = log2 ﹐
m m +1 m + 2 m + n −1 m
m+n
∴ n= ⇒ mn = m + n ⇒ mn − m − n = 0 ⇒ (m − 1) (n − 1) = 1﹐
m
－ 2 －

3. ⎧ m − 1 = 1 ⎧ m − 1 = −1 ⎧m = 2 ⎧m = 0
∴ ⎨ 或⎨ （∵ m﹐n ∈ N） ⇒ ⎨ 或⎨ （不合）
⎩ n − 1 = 1 ⎩ n − 1 = −1 ⎩n = 2 ⎩n = 0
⇒ 2m − 3n = 4 − 6 = −2﹒
1 1 log a 2 log b 2 log c 2
11.設 a = ﹐b = 2 ﹐c = 4 且 = + + ﹐則 P = ____________﹒
2 P log 2 b log 2 c log 2 a
1
解答 −
4
1 1
解析 loga2 = log 1 2 = log 1 2= = −2﹐∴ log2a = − ﹐
−
1 2
−
2 2
2
2
1
logb2 = log 2 2 = log24 = 2﹐∴ log2b = ﹐
2
1
logc2 = log42 = log 22 2 = ﹐∴ log2c = 2
2
1
1 −2 2 1
⇒ = + + 2 = −4 + 1 − 1 = −4﹐∴ P = − ﹒
P 1 2 −1 4
2 2
﹒
12.下表是函數 f ( x) = b + log a x 的四個函數值：
x 0.25 2 4 8
f (x) n m 10 − n m+4
試求 a﹐b 的值﹒
解答 a = 2 ﹐b = 5
⎧ 1
⎪n = b + log a ……(1) ⎧m = b + log a 2 ……(2)
解析 依序代入得 ⎨ 4 ⎨
⎪10 − n = b + log a 4 ……(3) ⎩m + 4 = b + log a 8 ……(4)
⎩
⎧n = b − 2log a 2 ⎧m = b + log a 2
⎨ ﹐⎨ ﹐得 b = 5 ﹐ a = 2 ﹒
⎩10 − n = b + 2log a 2 ⎩m + 4 = b + 3log a 2
－ 3 －