Poor y

Since $x, y$ and $z$ are consecutive natural numbers, then $x=y-1, z=y+1$ .

$1+xz$ , therefore, is equal to $1+(y-1)(y+1) \Rightarrow 1+y^2-1 \Rightarrow y^2$

$log(1+xz)=log(y^2)=\boxed{2log(y)}$

log(1+xz)=log(1+(y-1)(y+1))=log(y*y)=2log(y) : x,y,z consecutive natural numbers

The numbers are x, y = x+1 and z = y+1 = x+2 $$ $$ now, $$ log(1+xz) $$ = log{1+x(x+2)} $$ = log(1+x²+2x) $$ = log(1²+x²+2.x.1) $$ = log(1+x)² $$ = 2 log(1+x) $$ = 2logy

2logy is solution

Since it didn't specify values, we can take any values we like. Let x = 9, y = 10, z= 11. Then log (1 + xz) = log (100) = 2 and log y = log 10 = 1. 2 = 2(1).