Lost Logarithm III

Taking the product of the equations we get

$x^2y^2z^2=10^{a+b+c}$ .

Hence taking logs of both sides,

$\log (x^2y^2z^2)=a+b+c$

or $2\log (xyz)=a+b+c$

or $\log x+\log y+\log z=\dfrac{a+b+c}{2}$

Given, xy=10^a

log(xy)=log10^a (Taking log of each side)

log x + log y= a .............(1)

So we can say respectively,

log y + log z= b ..........(2)

log z + log x= c ...........(3)

(1)+(2)+(3) we get,

log x + logy + logy + log z + log z +log x =a+b+c

2(log x + logy + log z) = a+b+c

log x + logy + log z = (a+b+c)/2

(xy)(yz)(xz) = 10^(a + b + c)

(xyz)^2 = 10^(a + b + c)

2log(xyz) = a + b + c

log(xyz) = (a + b + c)/2

log x + log y + log z = ( a + b + c)/2

(x^2) (y^2) (z^2)=10^a+b+c

log(x^2) (y^2) (z^2)=a+b+c

2log(xyz)=a+b+c

log(xyz)=(a+b+c)/2

log x+log y+log z=(a+b+c)/2

log x+log y+log y+log z+log z+log x= a log10+b log10+c log10. or,2(log x+log y+log z)=a+b+c. or,log x+log y+log z=(a+b+c)/2

logx+logy=a logy+logz=b logx+logz=c adding these 3 2logx+2logy+2logz=a+b+c 2(logx+logy+logz)=a+b+c logx+logy+logz=a+b+c/2

log x+log y = a log 10 .........(1) log y+log z= b log 10 .........(2) log x+log z = c log 10 .........(3) summing (1).(2).(3) 2 log x+ 2 log y+ 2 log z= a + b + c dividing by 2 log x+ log y+ log z= (a + b + c)/2