由log[ log( logx)] = 0得，log( logx)= 1，logx =，即x = 2；
由log[ log( logy)] = 0得，log( logy) = 1，logy =，即y =3；由log[ log( logz)] = 0得，log( logz) = 1，logz =，即z = 5．
∵y =3= 3= 9，∴x = 2= 2= 8，∴y＞x，又∵x = 2= 2= 32，z = 5= 5= 25，∴x＞z．
故y＞x＞z．
请在这里输入关键词：
科目：高中数学
已知an=log（n+2）（n+3），我们把使乘积a1•a2•a3•…•an为整数的数n称为“优数”，则在区间（0，2012）内所有优数的个数为（　　）A．3B．4C．5D．6
科目：高中数学
已知log(2x+5)(x2+x-1)=1，则x的值是（　　）A、-4B、-2或3C、3D、-4或5
科目：高中数学
已知log[ log( logx)] = log[ log( logy)] = log[ log( logz)] = 0，试比较x、y、z的大小．
科目：高中数学
来源：2012年人教A版高中数学必修1对数函数练习卷（解析版）
题型：解答题
已知log[ log( logx)] = log[ log( logy)] = log[ log( logz)] = 0，试比较x、y、z的大小．
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我国年的财政收入和国民生产总值的计量分析_经济学毕业论文
我国年的财政收入和国民生产总值的计量分析
1978年十一届三中全会确立了改革开放的战略决策，在这一战略决策的指引下，我国的国民经济得到了飞速的发展,我国的总体经济实力不断增强，国民生产总值持续增长，总量已经位居世界前列，我国已经在经济发展上取得了举世瞩目的成就。
随着国民生产总值的增长，我国的财政收入也呈每年增长的趋势：
一.建立模型
我们知道国民生产总值是影响财政收入的主要因素，国民生产总值X与财政收入Y之间存在密切的关系，财政收入随国民生产总值的增加而增加，但变动幅度相对较低，因此可设定财政收入Y与国民生产总值X之间的关系为
Y=β1+β2X+U
其中：Y为年财政收入（亿元）；Xi为年国民生产总值（亿元）。变量采用年度数据，样本期为年。β1指当国民生产总值为零时的最低财政收入。
二．估计模型中的未知参数
假定模型中的随机误差项U满足古典假设，运用OLS方法估计模型的参数。
1．建立文档，输入数据
2．用OLS估计未知参数
所以模型是 Y=858.031X+U
三 模型检验
从估计结果、可以看出，可决系数为0.9916，表面上看模型在拟合上非常好。系数显著性检验：对于β2，t统计量为34.41495。给定α=0.05，查t分布表，在自由度为n-2=18下，。。。。。。得临界值t 0.025(18)=2.101,因为tt 0.025(18),所以拒绝H。：β2=0,
表明国民生产总值对财政收入有显著影响。并且从经济意义上看， =0.100031,表面国民生产总值每增加1亿元，财政收入平均增加0.11亿元。
若1998年的国民生产总值为78017.8亿元，下面预测1998年的财政收入。
通过Eviews计算得Y=元。通过下图可以看到
五，自相关性检验
根据上述OLS估计。我们暂把Y=858.031X+U
作为模型。根据其得到残差resid,运用genr生成序列e，则在quick菜单中选graph项，在图形对话框里键入：e e(-1)，再点击scatter diogram。得输出结果：
从上图可以看出残差E呈线性自回归，表明随机误差U存在自相关
根据Yi=858.031Xi+Ui的估计结果。由DW=0.8595,给定显著性水平α=0.05，查Durbin-Watson表，n=20,k’(解释变量个数)=1,得下限临界值d =1.20,上限临界值d=1.41,因为DW统计量为0.8595dL=1.20。根据判断区域可知，这时随机误差项存在正的一阶自相关。
六.自相关的修正
1.由DW=0.8595,根据 ρ=1-DW/2，计算出 ρ=0.57025。用GENR分别对X和Y作广义差分。在Workfile框中选GENR菜单，在对话框中直接输入生成格式。即
DX=X-0.57025*X(-1)
DY=Y-0.57025*Y(-1)
然后再用OLS方法估计其参数，结果为
DY=363.975DX
t=(5.938618) (27.40193)
可决系数为0.977861, F=750.8658, DW=1.076900.
这时，我们发现用广义差分法后，DW值有显著提高，但仍然存在自相关。
在Quick菜单中选Estimate Equation项，出现估计对话框，直接键入：Y C X AR(1),得如下结果：见下图:
从上表可以看到DW=1.087370,继续有所提高，但仍存在自相关性。
3.利用对数线性回归修正自相关。运用GENR分别对X和Y生成logX和logY，即
GENR LY=LOG(Y)
GENR LX=LOG(X)
在估计对话框中直接键入：LY C LX，即得相应的输出结果，见下表：
DW值不升反降，
再同时考虑迭代，在估计对话框里键入：LY C LX AR(1) 得下表：
故可知该模型始终存在自相关性。我们估计这是由于政府在制定财政预算时，总要根据上年甚至前几年的财政收入数据进行编制。这就是说本年的财政收入要受前几年的财政收入的影响。
七.异方差性检验
先将时间定为，然后用OLS方法得到下列结果
Y=535.279X
R^2=0.885479 ∑e1^2=23658.52
然后将时间定义为1991——1997，再用OLS方法得到如下结果
Y=773.069X
(2.507852) (16.48025)
R^2=0.981923 ∑e2^2=
求F统计量： F=/.19，查F分布表，给定显著性水平α=0.05，得临界值F0.05(6,6)=4.28,比较f=19.194.28,则拒绝原假设，表明随机误差存在异方差。
八.异方差的修正
WLS估计法。首先生成权数W=1/X，然后用OLS估计得出结果：
对数变换法
先用GENR生成序列LX和LY
GENR LY=LOG(Y)
GENR LX=LOG(X)
然后用OLS法求LY对LX的回归，其结果为：
比较第一中方法与第二种方法，我们发现X与Y在对数线形回归下拟合效果最好。这与财政收入Y的曲线呈对数型图形有关。
lnY=1..662427lnX
可决系数为0.989401 F统计量为
九，建立自回归模型
假设模型为：Y=α+β0*Xt+β1*Yt-1+Ut
回归结果如下：
结果显示， t检验值，F检验值，及R^2都显著，且h统计量 h=1.142
在显著水平α=0.05上，查标准分布表得临界值Hα/2=1.96,由于|h |=1.142〈1.96
则接受原假设ρ=0，模型扰动项不存在一介序列相关。估计模型为
Y=182.672Xt+0.7714Yt-1
(1.81) (4.197) (7.289)
R^2=0.997987 F= DW=1.55
该模型较好的解释了国明生产总值和财政收入之间的关系。
表明当年的财政收入不仅取决于当年的国民生产总值，还和上一年的财政收入有关。
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菁优解析考点：；．专题：综合题；不等式的解法及应用．分析：由题意，xyz（x+y+z）=1，1展开（x+y）（y+z），利用已知条件，构造基本不等式，求出最小值即可．解答：解：∵lgx+lgy+lgz+lg（x+y+z）=0，∴lg[xyz（x+y+z）]=0，∴xyz（x+y+z）=1，∴（x+y）（y+z）=xy+y2+yz+zx=y（x+y+z）+zx≥2=2．（当且仅当y（x+y+z）=zx时取等号）∴log2（x+y）+log2（y+z）=log2[（x+y）（y+z）]≥1，∴log2（x+y）+log2（y+z）的最小值为1故答案为：1．点评：本题是中档题，考查基本不等式求表达式的最小值问题，构造基本不等式是本题解题的关键，注意基本不等式满足的条件．答题：刘长柏老师　
其它回答（1条）
lgx+lgy+lg（x+y+z）=0& 即&& xyz（x+y+z）=1；& 而（x+y）（y+z）=xy+y^2+yz+xz=（x+y+z）y+xz≥2（（x+y+z）xz）^1/2 =2&& ∴log2（x+y）+log2（y+z）≤log=1&& 最小值为1
&&&&，V2.17943Math Forum - Ask Dr. Math
Solving the Equation x^y = y^x
Date: 12/09/2004 at 15:33:07
From: Chuck
Subject: Cannot find rigorous solution, just the obvious.
I cannot find a rigorous solution to the following:
Solve for X in terms of Y only:
(X to the Y power) = (Y to
the X power)
I can see obvious partial solutions like X = Y, but cannot derive
other solutions like 2,4 and 4,2.
Date: 12/09/2004 at 16:28:21
From: Doctor Vogler
Subject: Re: Cannot find rigorous solution, just the obvious.
Thanks for writing to Dr. Math.
In fact, this is a transcendental
equation, so you can't write the solution in the form
for any simple function (closed-form) f.
But let me tell you how you
would be best advised to analyze this equation:
First take the log of both sides:
log(X^Y) = log(Y^X)
and simplify:
Y*log(X) = X*log(Y)
and then divide by X*Y:
------ = ------.
Now you should consider the function
f(x) = ------.
Clearly, we have a solution to the last equation if and only if
f(X) = f(Y).
Well, this happens when X = Y, but does it happen elsewhere?
If we graph
we will find that f increases from y = -infinity at x = 0 to y = 1/e
at x = e (that's e = 2.71828... whether you used the common log or the
natural log or the log to any other base), and then f decreases from
y = 1/e at x = e to y = 0 at x = infinity.
Well, if X and Y are different values and
f(X) = f(Y),
then that means that there is a horizontal line which passes through
our function at two points (namely X and Y).
Look at the function,
and you'll find that the smaller value is somewhere between 1 and e,
and the larger value is bigger than e.
Also, the closer the smaller
value is to e, the closer the larger value is to e.
The closer the
smaller value is to 1, the bigger the larger value is.
So what you find is that if X &= 1, then the only solution is Y = X.
Similarly, if X = e, then the only solution is Y = X.
1 & X & e,
then there are exactly two solutions for Y, one of which is Y = X, and
the other is some number bigger than e.
Similarly, if
then there are exactly two solutions for Y, one of which is Y = X, and
the other is some number between 1 and e.
But can you write out a formula for the smaller value in terms of the
bigger value, or vice-versa?
Well, not using any closed-form
But you can use numerical methods to find approximate
solutions for any X value.
If you have any questions about this or need more help, please write
back and show me what you have been able to do, and I will try to
offer further suggestions.
- Doctor Vogler, The Math Forum
Date: 12/15/2004 at 14:39:28
From: Doctor Vogler
Subject: Re: Cannot find rigorous solution, just the obvious.
Hi again, Chuck,
It occurs to me that there is something else about this equation that
you might be interested in.
The solutions you gave (2, 4) and (4, 2)
are in integers.
In fact, these solutions and X = Y are the only
solutions in positive integers.
And the only integer solutions are X
= Y and (2, 4), (4, 2), (-2,-4), (-4, -2).
Proving that is as follows:
First suppose that X and Y are positive.
By switching the order of X and Y, we may assume that Y &= X.
divide both sides of the equation by X^X.
X^(Y-X) = (Y/X)^X.
Since the left side is clearly an integer, the right side has to be an
But if you raise a non-integer rational number to an integer
power, then you don't get an integer.
So that means that
must be an integer (bigger than 0).
Now we re-write our equation as
X^(kX - X) = k^X.
We take the positive real X'th root of both sides of the equation and get
X^(k-1) = k.
Now if X &= 2, then:
(a) k = 1 always works (and means X = Y)
(b) k = 2 implies X = X^(2-1) = 2 (and gives your solutions)
(c) k = 3 implies X^(k-1) & k
(d) by induction on k, k &= 3 implies
X^(k-1) = X*X^(k-2) & X(k-1) &= 2k-2 & k,
so there are only the solutions already mentioned when X &= 2.
But X = 1 implies Y = 1.
And X = 0 implies Y = 0.
And if X is
negative, but Y is positive, then Y^X is positive, so X^Y is positive,
which means that Y is even and
X^Y = (-X)^Y = Y^X,
(-X)^Y * Y^(-X) = 1,
but then (-X)^Y and Y^(-X) both have to be 1, so X is -1 and Y = 1.
Finally, if X and Y are both negative, then we raise both sides to the
-1 power and get
X^(-Y) = Y^(-X)
and then if X is odd and Y is even or vice-versa, then the signs don't
match, but if X and Y are both odd, then we multiply both sides of the
equation by -1 to get
(-X)^(-Y) = (-Y)^(-X).
If both X and Y are even, then we don't need to multiply, and we still
get the same equation.
So (-X, -Y) is a solution in positive integers.
Again, if you have any questions about this or need more help, please
write back and show me what you have been able to do, and I will try
to offer further suggestions.
- Doctor Vogler, The Math Forum
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http://mathforum.org/dr.math/函数若lgx+lgy=1时,则x分之5+y分之2的最小值, 函数若lgx+lgy=1时,则x分之5+y
函数若lgx+lgy=1时,则x分之5+y分之2的最小值
完美萨克斯风 函数若lgx+lgy=1时,则x分之5+y分之2的最小值
lg(xy)=1xy=105/x+2/y=(5y+2x)/xy=(5y+2x)/10真数x&0,y&05y+2x&=2√(5y骸酣汾叫莴既风习袱卢*2x)=2√(10*10)=20当5y=2x时取等号y=2x/5,xy=10,有正数解所以等号能取到所以最小值=20/10=2
将上式：lgx+lgy=1，进行变换得到：lg(xy)=1，即xy=10因为5/x+2/y=(5y+2x)/xy=(5y+2x)/10真数x&0,y&05y+2x&=2√(5y*2x)=2√(10*10)=20当5y=2x时取等号y=2x/5,xy骸酣汾叫莴既风习袱卢=10,有正数解所以等号能取到从而最小值为20/10=2}