# solve: (yz)^log y-log z.(zx)^log z-logx .(xy)^log x -logy                                      solve: (yz)^log y-log z.(zx)^log z-logx .(xy)^log x -logy                ...

solve: (yz)^log y-log z.(zx)^log z-logx .(xy)^log x -logy

solve: (yz)^log y-log z.(zx)^log z-logx .(xy)^log x -logy

the dot represents multiplication sign

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

Supposing that you want to solve the expression `(yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy), ` you need to take logarithm of this expression such that

`log((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy)) = log((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy))` Converting the logarithm of the product into a sum of logarithms yields:

`log((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy)) = log((yz)^(log y-log z)) + log ((zx)^(log z-logx))+ log ((xy)^(log x -logy))` Using the logarithmic identities yields:

`log((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy)) = (log y-log z)(log (yz)) + (log z-logx)(log(xz)) + (log x -logy)(log(xy))`

You need to convert the logarithm of product into the sum of logarithms such that:

`log((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy)) = (log y - log z)(log y + log z) + (log z - log x)(log z+ log x) + (log x - log y)(log x +log y)`

You should convert the special products into differences of squares such that:

`log((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy)) = log^2y - log^2 z + log^2 z - log^2 x + log^2 x - log^2 y`

Reducing like terms yields:

`log((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy)) = 0 =gt ((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy))  = 10^0`

`((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy)) = 1`

Hence, evaluating the value of given expression yields `((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy)) = 1.`

Sources: