高職數學log題

1.設實數x滿足logx_4－log2_x=1，則x=
Sol
logx_4－log2_x=1
log4/logx －logx/log2=1
設 p=log2/logx
2p－1/p=1
2p^2－p－1=0
(2p+1)(p－1)=0
(1) 2p+1=0
p=－1/2
log2/logx=－1/2
logx=－2log2=log(1/4)
x=1/4
(2) p－1=0
p=1
log2/logx=1
logx=log2
x=2

2．(log2_3+log4_9)*(log3_16+log9_2)=
Sol
(log2_3+log4_9)*(log3_16+log9_2)
=(log3/log2+log9/log4)*(log16/log3+log2/log9)
=(log3/log2+log3/log2)*[4log2/log3+(1/2)log2/log3]
=(2log3/log2)*(9/2)*(log2/log3)
=9

3.設a，b，c為正整數，若alog150_2+blog150_3+clog150_5=4則a+b－c=?
Sol
alog150_2+blog150_3+clog150_5=4
alog2/log150+blog3/log150+clog5/log150=4
alog2+blog3+clog5=4log150
(2^a)*(3\^b)*(5^c)=150^4
(2^a)*(3\^b)*(5^c)=(2^4)*(3^4)*(5^8)
a=4，b=4，c=8
a+b－c=0

第一題
令A=LOGX, B=LOG2
原式化為2B/A -A/B =1
(2B^2 -A^2 -AB)/AB =0
故2B^2 -A^2 -AB=0, AB/=0
A^2 +AB -2B^2=0, A=B或-2B
意即LOGX=LOG2或-2LOG2
ANS：2或1/4


第二題
一樣的方法，令A=LOG2, B=LOG3
原式化為(B/A +2B/2A)(4A/B +A/2B)
= [(1+2/2)(B/A)][(4/1 +1/2)(A/B)]
= 2*(9/2)
ANS：9


第三題
LOG(150,2^A) +LOG(150,3^B) +LOG(150,5^C) =LOG(150,150^4)
2^A *3^B *5^C = (2*3*5^2)^4
A+B-C=4+4-8
ANS：0

1. Ans: x = 2 or 1/4

1.設實數x滿足logx ４－log2 Ｘ＝１,則X＝
設 z=log x, a=log 2
2a/z - z/a=1
2a^2-z^2=az
z^2+az-2a^2=0
(z-a)(z+2a)=0
z=a or -2a (不合)
log x = log 2
x=2 ...Ans

２．（log2 ３＋log4 ９）（log3 １６＋log9 ２)=
設 a=log 2, b=log 3
原式 = [b/a+(2b)/(2a)][4a/b+a/(2b)]=9 ...Ans


3.設a,b,c為正整數,若a log150 ２＋b log150３＋c log150５＝４則ａ＋ｂ－ｃ＝
(2^a)(3^b)(5^c)=150^4=(2*3*5^2)^4=(2^4)(3^4)(5^8)
a=4, b=4, c=8
a+b-c=0 ...Ans