解:(1)設(shè)t=
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,任取x
2<x
1<-5,則
t
2-t
1=
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-
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=
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=
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.
∵x
1<-5,x
2<-5,x
2<x
1,
∴x
1+5<0,x
2+5<0,x
2-x
1<0.
∴
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<0,即t
2<t
1.
當(dāng)a>1時(shí),y=log
ax是增函數(shù),∴l(xiāng)og
at
2<log
at
1,即f(x
2)<f(x
1);
當(dāng)0<a<1時(shí),y=log
ax是減函數(shù),∴l(xiāng)og
at
2>log
at
1,即f(x
2)>f(x
1).
綜上可知,當(dāng)a>1時(shí),f(x)在區(qū)間(-∞,-5)為增函數(shù);
當(dāng)0<a<1時(shí),f(x)在區(qū)間(-∞,-5)為減函數(shù).
(2)g(x)=1+log
a(x-3)=log
aa(x-3),
方程f(x)=g(x)等價(jià)于:
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即方程
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在區(qū)間(5,+∞)上有解,
∵
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=
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∴函數(shù)F(x)=
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在區(qū)間(5,5+2
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)上導(dǎo)數(shù)大于零,在區(qū)間(5+2
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,+∞)導(dǎo)數(shù)小于零
可得F(x)=
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在區(qū)間(5,5+2
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)上單調(diào)增,在區(qū)間(5+2
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,+∞)單調(diào)減
∴F(x)的最大值為F(5+2
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)=
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,而F(x)的最小值大于F(5)=0
要使方程方程
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在區(qū)間(5,+∞)上有解,必須a∈(0,
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)
所以a的取值范圍是:(0,
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)
分析:(1)將對數(shù)的真數(shù)當(dāng)成一個(gè)函數(shù),可以用定義證明它在區(qū)間(-∞,-5)內(nèi)的單調(diào)性,再討論底數(shù)a與1的大小關(guān)系得到相應(yīng)的情況下真數(shù)的大小關(guān)系,即可得函數(shù)f(x)在區(qū)間(-∞,-5)內(nèi)的單調(diào)性;
(2)化函數(shù)g(x)=1+log
a(x-3))為g(x)=log
aa(x-3),方程f(x)=g(x)即為它們的真數(shù)都大于零且相等,采用變量分離的方法,轉(zhuǎn)化為求函數(shù)F(x)=
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在區(qū)間(5,+∞)上的值域,實(shí)數(shù)a的取值范圍就應(yīng)該屬于這個(gè)值域.
點(diǎn)評:本題著重考查了函數(shù)單調(diào)性的判斷與證明、根的存在性及根的個(gè)數(shù)判斷等知識點(diǎn),在解題時(shí)應(yīng)該注意分類討論與數(shù)形結(jié)合等數(shù)學(xué)思想的應(yīng)用.