Question 152980
Three rules to use here.
{{{log(b,X)+log(b,Y)=log(b,XY)}}}
{{{log(b,X)-log(b,Y)=log(b,(X/Y))}}}
{{{a*log(b,Z)=log(b,Z^a)}}}
I'm not sure of your equation but here's what I think it is,
1.{{{3*log(b,x)-4*log(b,y)+(1/2)log(b,z)}}}
although it might be,
2.{{{3*log(b,(x-4))-log(b,y)+(1/2)log(b,z)}}}
Let's look at each term separately, then we'll put it all together.
{{{3*log(b,x)=log(b,x^3)}}}
{{{4*log(b,y)=log(b,y^4)}}}
{{{(1/2)log(b,z)=log(b,z^(1/2))}}}
{{{3*log(b,x)-4*log(b,y)+(1/2)log(b,z)=log(b,x^3)-log(b,y^4)+log(b,z^(1/2))}}}
{{{3*log(b,x)-4*log(b,y)+(1/2)log(b,z)=log(b,x^3/y^4)+log(b,z^(1/2))}}}
{{{3*log(b,x)-4*log(b,y)+(1/2)log(b,z)=log(b,((x^3*z^(1/2))/y^4))}}}
If it's eq. 2, then the solution becomes,
{{{3*log(b,(x-4))-log(b,y)+(1/2)log(b,z)=log(b,(((x-4)^3*z^(1/2))/y))}}}