Question

if
X= 1+loga bc
Y=1+ logb ca
Z=1= logc ab
solve XYZ

plz

Answer 1

you mean, if x = 1 + loga (bc), y = 1 + logb (ca) and z = 1 + logc (ca)

then we have to find xyz

solution : we know from logarithm properties

• logr (mn) = logr m + logr n
• and logr (m/n) = logr m - logr n
• $log_rr$ = 1
• $log_mn=\frac{1}{log_nm}$

now, x = 1 + loga (bc)

= loga (a) + loga (bc)

= loga (abc)

or, 1/x = log(abc) a .....(1)

similarly, y = 1 + logb (ca) = logb (abc)

or, 1/y = log(abc) b ......(2)

and z = 1 + logc (ab) = logc (abc)

or, 1/z = log(abc) c ......(3)

now, 1/x + 1/y + 1/z = log(abc) a + log(abc) b + log(abc) c

= log(abc) (a × b × c)

= log(abc) (abc)

= 1

or, 1/x + 1/y + 1/z = 1

or, (yz + yz + xy)/xyz = 1

or, xyz = xy + yz + zx

hence, xyz = xy + yz + zx

Answer 2

Answer:

xy + yz +zx =xyz

Step-by-step explanation:

x = 1+ log a bc =loga a+ loga bc

= loga abc

therefore 1/x=logabc a

similarly y = log b abc

z= log c abc

(xy +yz+ zx)/xyz

= 1/z + 1/x +1/y

= logabc c +logabc a +logabc b

= log abc abc

=1

(xy +yz+ zx)/xyz = 1

therefore xy + yz +zx =xyz