solve..
8^(x+2) = 3[4^(2x)]
math logarithm question..
2008-10-15 9:39 pm
回答 (2)
2008-10-15 9:51 pm
✔ 最佳答案
8^(x + 2) = 3[4^(2x)][2^3]^(x +2) = 3[2^4x]
2^(3x + 6) = 3[2^4x]
Let 2^x = y, so 2^3x = y^3 and 2^4x = y^4.
The equation becomes
64y^3 = 3y^4
y^3(3y - 64) = 0
y = 0 (rej.)
y = 64/3, that is
2^x = 64/3
x = log(64/3)/log2 = 1.329/0.301= 4.415.
2008-10-16 12:37 am
You can try to use log first.
log[ 8^(x+2)] = log {3[4^(2x)]}
(x+2) log 8 = log 3 + (2x) log 4 (log ab = log a + log b)
(3x+6) log 2 - 4x log 2 = log 3
6 log 2 - x log2 =log 3
x=6 - 1og 3 / log 2
2008-10-15 16:38:26 補充：
Sorry, typo error
x=6 - log 3 / log 2
log[ 8^(x+2)] = log {3[4^(2x)]}
(x+2) log 8 = log 3 + (2x) log 4 (log ab = log a + log b)
(3x+6) log 2 - 4x log 2 = log 3
6 log 2 - x log2 =log 3
x=6 - 1og 3 / log 2
2008-10-15 16:38:26 補充：
Sorry, typo error
x=6 - log 3 / log 2
收錄日期: 2021-04-25 22:35:38
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https://hk.answers.yahoo.com/question/index?qid=20081015000051KK00681