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第 13 週 · 指數與對數 · 第 5 節

解指數方程與對數方程

未知數躲在指數上（2ˣ = 8）或躲在對數裡（log x = 1）怎麼解？四招拆穿它 —— 同底比指數、化同底、換元變二次、對數化指數，外加一條鐵律：解完必驗真數。配滿圖例。

▶ 第一關： 同底比指數
same base → compare the exponentsax=akbases match → cancel⟹x=ke.g. 2ˣ = 2³ ⟹ x = 3
2ˣ = 8→2ˣ = 2³→x = 3rewrite 8 = 2³same base 2compare!
memorize these powers8=2³ 16=2⁴ 32=2⁵9=3² 27=3³ 81=3⁴25=5² 125=5³rewrite the right side, then compare exponents
▶ 第二關： 化成同底
4ˣ = 8→(2²)ˣ = 2³→2²ˣ = 2³→2x = 3→x = 3/24 = 2², so (2²)ˣ = 2^(2x)both base 2 → compare
fraction base → flip up with a minus sign(1/2)ˣ = 8→2⁻ˣ = 2³→x = −31/2 = 2⁻¹−x = 3negative is OK
▶ 第三關： 換元變二次 ＋ 對數化指數
substitute t = 2ˣ (so 4ˣ = t²)4ˣ − 3·2ˣ − 4 = 0→t² − 3t − 4 = 0→(t−4)(t+1)=0→t=4t=−1 ✗ (t>0)→2ˣ = 4 = 2²→x = 2
log and exponent are inverses (W13D2)logₐ x = k⟺x = aᵏlog₂ x = 5 → x = 2⁵ = 32
⚔ BOSS： 合併、換元與驗真數
merge logs → solve → CHECK argumentslog x + log(x−3) = 1→log[x(x−3)] = 1→x(x−3) = 10→x²−3x−10=0→(x−5)(x+2)=0x=5 ✓x=−2 ✗ arg<0x=−2 makes log(−2) undefined → reject
the argument N of logₐ N must be > 00N ≤ 0 → log undefined ✗N > 0 → allowed ✓any root landing on the red side is rejected
BOSS worked examplelog₂(x+1)+log₂(x−1)=3→(x+1)(x−1)=2³=8→x²=9→x=3 or x=−3x=3 ✓x=−3 ✗ x+1<0check each argument > 0 before you commit