log4a=log16b=log 64(a+b),求a之值

Question

設a,b為正實數且log4a=log16b=log 64(a+b),求a之值

註:4,16及64為底數,a及b為真數

Answer 1

最佳解答

log(4) a = log(16) b = log(64) (a + b)

==> log a / log 4 = log b / log 16 = log (a + b) / log 64

==> log a / log 4 = log b / log 16 = log (a + b) / log 64

所以

log a / log 4 ＝ log b / log 16

==> log a / log 4 ＝ log b / 2 log 4

即 a^2 = b ⋯⋯⋯⋯⋯⋯ (i)

log a / log 4 = log (a + b) / log 64

==> log a / log 4 = log (a + b) / 3 log 4

即 a^3 = a + b

==> a^3 = a + a^2 ⋯⋯ (由 (i))

==> a^3 - a^2 - a = 0

==> a(a^2 - a - 1) = 0

==> a = 0 (捨去) 或 a＝(1＋√5)/2 或 a＝(1－√5)/2 (捨去)

答案：a＝(1＋√5)/2

(代入 (i), 得 b＝(3＋√5)/2)