1已知解方程式㏒2(x+a)=㏒4(11x+b)可得x=2或x=3，求數對(a，b).其中未括號的2和4為底數.?

先留意真數需大於零.即 x>-a且x>(-b/11)
已知方程式log₂(x+a)=log₄(11x+b)
=>log₂(x+a)=(1/2)*log₂(11x+b)
=>log₂(x + a)²=log₂(11x+b)皆以2為底數
分別用x=2和x=3代入得兩條方程式:
a^2 + 4a + 4 = 22 + b 和 a^2 + 6a + 9 = 33 + b
解得 a=3和b=3 符合真數需大於零的前提.即 x>-a且x>(-b/11)
所以數對 (a，b)=(3，3)

log₂(x + a) = log₄(11x + b)
log(x + a) / log(2) = log(11x + b) / log(4)
log(x + a) / log(2) = log(11x + b) / log(2²)
log(x + a) / log(2) = log(11x + b) / [2 log(2)]
log(x + a) = log(11x + b) / 2
2 log(x + a) = log(11x + b)
log(x + a)² = log(11x + b)
(x + a)² = 11x + b ...... [1]

把 x = 2 代入 [1] 中：
(2 + a)² = 11(2) + b
a² + 4a + 4 = 22 + b
a² + 4a - 18 = b ...... [2]

把 x = 3 代入 [1] 中：
(3 + a)² = 11(3) + b
a² + 6a + 9 = 33 + b
a² + 6a - 24 = b ...... [3]

[3] - [2] :
2a - 6 = 0
a = 3

把 a = 3 代入 [2] 中：
3² + 4(3) - 18 = b
b = 3

所以 (a, b) = (3, 3)

[log(x+a)]/(log 2)=[log (11x+b)]/(log 4)
2 log(x+a)=log(11x+b)
x^2 +2ax+a^2=11x+b
x^2 +(2a-11)x+(a^2-b)=0
4+2(2a-11)+(a^2-b)=0
a^2+4a-b=18 i
9+3(2a-11)+(a^2-b)=0
a^2 +6a-b=24 ii
ii-i
2a=6
a=3
代a=3入i
9+4(9)-b=18
b=27
(a,b)=(3,27)