fun1=@(x,y,z)log((x^2+y^2+z^2);g=triplequad(fun1,90,110,50,70,20,40)Error using ==> mpowerInputs must be a scalar and a square matrix.Error in ==> @(x,y,z)log(x^2+y^2+z^2)Error in ==> quad at 77y = f(x,varargin{:});Error in ==> triplequad>innerintegr

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fun1=@(x,y,z)log((x^2+y^2+z^2);g=triplequad(fun1,90,110,50,70,20,40)Error using ==> mpowerInputs must be a scalar and a square matrix.Error in ==> @(x,y,z)log(x^2+y^2+z^2)Error in ==> quad at 77y = f(x,varargin{:});Error in ==> triplequad>innerintegr
fun1=@(x,y,z)log((x^2+y^2+z^2);g=triplequad(fun1,90,110,50,70,20,40)
Error using ==> mpowerInputs must be a scalar and a square matrix.
Error in ==> @(x,y,z)log(x^2+y^2+z^2)
Error in ==> quad at 77y = f(x,varargin{:});
Error in ==> triplequad>innerintegral at 69 Q(i) = quadf(intfcn,xmin,xmax,tol,trace,y(i),z,varargin{:});
Error in ==> dblquad>innerintegral at 80fcl = intfcn(xmin,y(1),varargin{:}); %evaluate only to get the class below
Error in ==> quad at 77y = f(x,varargin{:});
Error in ==> dblquad at 60Q = quadf(@innerintegral,ymin,ymax,tol,trace,intfcn,...
Error in ==> triplequad at 53Q = dblquad(@innerintegral,ymin,ymax,zmin,zmax,tol,quadf,intfcn,...

fun1=@(x,y,z)log((x^2+y^2+z^2);g=triplequad(fun1,90,110,50,70,20,40)Error using ==> mpowerInputs must be a scalar and a square matrix.Error in ==> @(x,y,z)log(x^2+y^2+z^2)Error in ==> quad at 77y = f(x,varargin{:});Error in ==> triplequad>innerintegr
矩阵运算要在运算符前加点的.

问题是什么啊

fun1=@(x,y,z)log((x^2+y^2+z^2);g=triplequad(fun1,90,110,50,70,20,40)Error using ==> mpowerInputs must be a scalar and a square matrix.Error in ==> @(x,y,z)log(x^2+y^2+z^2)Error in ==> quad at 77y = f(x,varargin{:});Error in ==> triplequad>innerintegr 2log(m)y-log(m)x-log(m)z 4.函数fun1的调用形式是:fun1(x,(y,z) ,10,fun2((x-1,y-1))),请问函数fun1到底有几个参数() A)1 B) [log(4)(2x)][log(2x)(3y)][log(3y)(z)]=log(8x)[(8x)]^2,则z=? log₂log₃log₄ X=log₃log₄log₂ Y=log₄log₂log₃ Z=0X+Y+Z=?为什么是71呀? 已知x>1,y>1,z>1,log(x)W=24,log(y)W=40,log(xyz)W=12求log(z)W log(2x)+log(3y)-log(2z) 怎么化简?log3(x^2-2x-6)=2 log(3x+6)=1+log(x)求解,求每步详...log(2x)+log(3y)-log(2z) 怎么化简?log3(x^2-2x-6)=2 log(3x+6)=1+log(x)求解, matlab 极小化函数f(x,y) = -9*y - 10*z + 100*(-1*log(100-y-z)) - log(y) - log(z) - log(50-y+z)matlab怎么极小化这个函数,我搞的老是出错 若log底数2[log底数3(log底数4)]若log底数2[log底数3(log底数4(x))]=log底数3[log底数4(log底数2(y))]=log底数4[log底数2(log底数3z)]=0,则x+y+z= log(2x)+log(3y)-log(2z) 怎么化简? log3(x^2-2x-6)=2 log(3x+6)=1+log(x)求解. 可以说log a x +log a y=log a 已知x,y,z均大于1,w不等于0,log x w=24,log y w=40,log xyz w=12,求log z w log(x*y+z)能不能找到一种方法能将其展开成log(x),log(y),log(z)之间的运算关系, 已知0<a<1,x=log a 根号2+log a 根号3y=1/2 log a 5z=log a 根号21 -log a 根号3比较x y z 的大小关系x=log a (根号2)+log a (根号3)y=1/2 (log a 5)z=log a (根号21) -log a (根号3) 怎么用matlab解这个方程组先是三个方程F(x,y,z)=log(((x^2+y^2+(z-5)^2)^0.5)-y)-log(((x^2+y^2+(z-5)^2)^0.5)+y);D(y,x,z)=log(((x^2+y^2+(z-5)^2)^0.5)-y)-log(((x^2+y^2+(z-5)^2)^0.5)+y); (第二个和第一个很类似,只是变量的位置 (x+y-z)(x-y+z)= 证明高一的换底公式你是用了换底公式不是证明它已经知道了:已知 A^B=C,设 A=X^Y,C=X^Z 代入,X^YB=X^Z ,所以 YB=Z,又,B=LOG(A,C),Y=LOG(X,A),Z=LOG(X,C) 故 LOG(A,C) = LOG(X,C) / LOG(X,A) X+Y+Z=?