maths 指數+對數

1.(b)
2^(3x - 2) = 6
log [2^(3x - 2)] = log 6
(3x - 2) log 2 = log 6
3x - 2 = log 6 / log 2
3x = [(log 6 / log 2) + 2]
x = [(log 6 / log 2) + 2] / 3
x ≈ 1.53 (3 sig. fig.)


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2.
log16 4(x + 1) - log4 8 = log16 (x - 2)
[log 4(x + 1) / log 16] - (log 8 / log 4) = [log (x - 2) / log 16]
[log 4(x + 1) / log 2^4] - (log 8 / log 2^2) = [log (x - 2) / log 2^4]
[log 4(x + 1) / 4 log 2] - (log 8 / 2 log 2) = [log (x - 2) / 4 log 2]
[log 4(x + 1) - 2 log 8] / 4 log 2 = log (x - 2) / 4 log 2
log 4(x + 1) - log 8^2 = log (x - 2)
log [4(x + 1) / 64] = log (x - 2)
4(x + 1) / 64 = x - 2
x + 1 = 16x - 32
15x = 33
x = 11/5
x = 2.2

1b
2^(3x-2)=6
2^(3x)=24
(3x)(log2)=log24
x=1.52

2
log[4(x+1)]/log16-log8/log4=log(x-2)/log16
log[4(x+1)/(x-2)]/log16=(3log2)/(2log2)
{log[4(x+1)/(x-2)]}/(4log2)=3/2
log[4(x+1)/(x-2)]=log64
4(x+1)/(x-2)=64
4(x+1)=64(x-2)
x+1=16x-32
33=15x
x=2.2

1b)
2^(3x-2) = 6
2^(3x)/4 = 6
8^x = 24
xlog8 = log24
x = log24/log8
x = 1.53 ##
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2)
留意 log{a} b = log b / log a, {a}即基數為a
另外, 從項log{16} (x-2) 得知x > 2
log{16} [4(x+1)] - log{4} 8 = log{16} (x-2)
[log (4x+4)] / log16 - [log 8]/[log4] = [log(x-2)] / log16
log (4x+4) - (log 8)(log16/log4) = log(x-2)

2012-06-25 23:06:21 補充：
(續)
log (4x+4) - (log8)(log{4}16) = log(x-2)
log (4x+4) - 2log8 = log (x-2)
log (4x+4) - log 8² = log (x-2)
log [(4x+4)/64] = log (x-2)
(x+1)/16 = x-2
15x = 33
x = 11/5 (>2)##