中四級-指數函數和對數函數

1. x^3 - y^3 = (x - y)(x^2 + xy + y^2)
log sqrt ( x^3 - y^3) = (1/2) log (x^3 - y^3) = (1/2) [ log (x- y)(x^2 + xy + y^2)
= (1/2) [ log (x - y) + log (x^2 + xy + y^2)] = (1/2) ( m + n).
2.
log 24 base 3 = log ( 3 x 2 x 2 x 2) base 3
= log 3 base 3 + log (2 x 2 x 2) base 3
= 1 + 3 log 2 base 3
= a
so log 2 base 3 = (a - 1)/3.
3.
a + b = 8
ab = 2.
(a)
log a base (a + b) + log b base (a + b)
= log ab base (a + b)
= log 2 base 8
= log 2 / log 8
= log 2 / (3 log 2) = 1/3.
(b)
log 9^a base 3 = log (3^2)^a base 3
= log 3^2a base 3
= 2a log 3 base 3
= 2a.
log ( 1/36^b) base 6
= log 36^(-b) base 6
= log 6^2(-b) base 6
= - 2b log 6 base 6
= - 2b
So their sum = 2a - 2b = 2(a -b).
(a - b)^2 = a^2 - 2ab + b^2 = (a + b)^2 - 4ab = 8^2 - 4(2) = 64 - 8 = 56
so 2(a - b) = 2sqrt 56 ( since a > b, take the +ve value).

第二題明顯是看錯題目。
log (8) 24 = a
==> log 24 = a log 8
==> log 3 + log 8 = a log 8
==> log 3 = (a - 1) log 8 = 2(a - 1) log 2
==> log 2 / log 3 = 1/[2(a - 1)]
==> log (3) 2 = 1/[2(a - 1)]

所以答案是 1/[2(a - 1)]

wy
1.正確
2.錯誤
3a正確
3b正確

2014-02-01 13:14:39 補充：
中間錯了少少~那些年~
log (8) 24 = a
==> log 24 = a log 8
==> log 3 + log 8 = a log 8
==> log 3 = (a - 1) log 8 = 3(a - 1) log 2
==> log 2 / log 3 = 1/[3(a - 1)]
==> log (3) 2
= 1/[3(a - 1)]

2014-02-01 13:30:57 補充：
大至上完全明白了，多謝大家的協助。