Step-by-step explanation:
Given:
log_(a) abc = x+1,
log_(b) abc = y+1,
log_(c) abc =z+1
Asked: x + y + z + 2 = ?
Solution:
Let
log_(a) abc = x+1 ⇛log_(abc) a = {1/(x+1)} →→→Eqn(1)
log_(b) abc = y+1 ⇛log_(abc) b = {1/(y+1)} →→→Eqn(2)
log_(c) abc = z+1 ⇛log_(abc) c = {1/(z+1)} →→→Eqn(3)
On adding equation (1),(2) and (3) then
log_(abc) a + log_(abc) b + log_(abc) c = {1/(x+1) + {1/(y+1)} + {1/(z+1)}
⇛log_(abc) abc = {{(y+1)(z+1) + (z+1)(x+1) + (x+1)(y+1)}/{(x+1)(y+1)(z+1)}]
⇛1 = {{(y+1)(z+1) + (z+1)(x+1) + (x+1)(y+1)}/{(x+1)(y+1)(z+1)}]
⇛{(x+1)(y+1)}(z+1) = (y+1)(z+1) + (z+1)(x+1) + (x+1)(y+1)
⇛{x(y+1)+1(y+1)}(z+1) = (y+1)(z+1) + (z+1)(x+1) + (x+1)(y+1)
⇛{xy + x + 1 + y}(z+1) = (y+1)(z+1) + (z+1)(x+1) + (x+1)(y+1)
⇛{xy + x + y + 1}(z+1) = (y+1)(z+1) + (z+1)(x+1) + (x+1)(y+1)
⇛z(xy + x + y + 1) + 1(xy + x + y + 1) = (y+1)(z+1) + (z+1)(x+1) + (x+1)(y+1)
⇛xyz + xz + yz + z + xy + x + y + 1 = (y+1)(z+1) + (z+1)(x+1) + (x+1)(y+1)
⇛xyz + xz + yz + z + xy + x + y + 1 = y(z+1)+1(z+1) + z(x+1)+1(x+1) + x(y+1)+1(y+1)
⇛xyz + xz + yz + z + xy + x + y + 1 = yz + y + z + 1 + xz + z + x + 1 + xy + x + y + 1
⇛x+y+z = y + z + 1 + x + 1
⇛x+y+z = x + y + z + 2
Therefore, x + y + z + 2 = x + y + z
Answer: Hence, the value of x + y + z + 2 = x + y + z.
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