高一的指數對數問題

1.
L(t) = a(1 - 10⁻ᵇᵗ)

L(1) = 78
a(1 - 10⁻ᵇ) =78 ...... [1]

L(3) = 143
a(1 - 10⁻²ᵇ) =143 ...... [2]

[1]/[2] :
(1 - 10⁻ᵇ)/(1- 10⁻²ᵇ) =78/143
78 - 78(10⁻²ᵇ) =143 - 143(10⁻ᵇ)
78(10⁻ᵇ)²- 143(10⁻ᵇ) +65 = 0
6(10⁻ᵇ)² - 11(10⁻ᵇ) + 5 = 0
[6(10⁻ᵇ) -5] [(10⁻ᵇ) -1] = 0
10⁻ᵇ = 5/6或 10⁻ᵇ =1
-b = log(5/6) 或 -b = log(1)
b = -log(5/6) 或 b = 0 (不合)

當 b= -log(5/6)：
log(10⁻ᵇ) =-b log(10)
log(10⁻ᵇ) =-b
log(10⁻ᵇ) =-[-log(5/6)]
log(10⁻ᵇ) =log(5/6)
10⁻ᵇ = 5/6...... [3]

把[3] 代入[1] :
a[1 - (5/6)] = 78
a = 468

L(t) = a(1 - 10⁻ᵇᵗ)
因此，L(t)= 468 [1 - (5/6)ᵗ]

L(3) = 468 [1 - (5/6)³]
L(3) = 197 (至整數)
三星期可以背熟的單字數目 = 197 個


2.
(5ˣ)² <5ˣ⌃²
5²ˣ <5ˣ⌃²
2x < x²
x² -2x > 0
x(x - 2) > 0
範圍： x < 0 或 x > 2


3.
log9(p) = log27(q) = log81(p + q)
log(p) / log(9) = log(q) / log(27) = log(p + q) / log(81)
log(p) / 2 log(3) = log(q) / 3 log(3) = log(p + q) / 4 log(3)
(log(p) / 2) × 12= (log(q) / 3) × 12= [log(p + q) / 4] × 12
6 log(p) = 4 log(q) = 3 log(p + q)

6 log(p) = 4 log(q)
3 log(p) = 2 log(q)
log(p³) = log(q²)
q = (√p)³ ...... [1]

6 log(p) = 3 log(p + q)
2 log(p) = log(p + q)
log(p²) =log(p + q)
p² = p+ q ...... [2]

把[1] 代入[2] 中：
p² = p+ (√p)³
(√p)⁴ - (√p)³ - (√p)² =0
p[(√p)² - (√p) - 1] = 0
p = 0 (不合) 或 √p =(1 + √5) / 2 或 √p =(1 - √5) / 2 (不合)

p = [(1 + √5) / 2]²
p = [(1 + 5 + 2√5)/ 4]
p = (3 + √5)/ 2


4.
設 u = log3(x)

[log3(x)]² < log3(x⁴)
[log3(x)]² < 4 log3(x)
u²< 4u
u² - 4u< 0
u(u - 4) < 0
0 < u < 4
0 < log3(x) < 4
log3(1) < log3(x) < 4 log3(3)
log3(1) < log3(x) < log3(3⁴)
log3(1) < log3(x) < log3(81)
1 < x < 81

第一題key：L(2)/L(1)
第三題：https://tw.knowledge.yahoo.com/question/question?qid=1014121301536