Question

If a^x-1=bc, b^y-1=ca, c^z-1=ab, then show that xy+yz+zx=xyz.​

Answer 1

given, a^(x - 1) = bc

b^(y - 1) = ca

c^(z - 1) = ab

we have to prove that xy + yz + zx = xyz

first of all, resolve, a^(x - 1) = bc ,b^(y - 1) = ca and c^(z - 1) = ab

a^(x - 1) = bc

⇒a^x . a^-1 = bc

⇒a^x/a = bc

⇒a^x = abc

⇒a = (abc)^{1/x}

similarly, b^(y - 1) = ca ⇒b = (abc)^{1/y}

and c^(z - 1) = ab ⇒c = (abc)^{1/z}

now, a.b.c = (abc)^{1/x}.(abc)^{1/y}. (abc)^{1/x}

taking log both sides,

log(abc) = log[ (abc)^{1/x}.(abc)^{1/y}. (abc)^{1/x}]

⇒log(abc) = log[(abc)^{1/x + 1/y + 1/z}]

⇒log(abc) = (1/x + 1/y + 1/z) log(abc)

⇒1 = 1/x + 1/y + 1/z

⇒1 = (yz + zx + xy)/xyz

⇒xyz = yz + zx + xy

hence proved

Answer 2

Step-by-step explanation:

Given that x = 1 + logₐ(bc)

Writing 1 as logₐ(a), we can rewrite x as

x = logₐ(a) + logₐ(bc)

Using Additive Property of logarithm

log m + log n = log mn, we get

x = logₐ(abc)

By Using change of base property

x = log abc/log a

So, 1/x = log a/log abc------(1)

Similarly as done for simplifying x,

Given y = 1 + log b(ca),

so we get 1/y = log b/log abc

Given z = 1 + log c(ab)

so we get 1/z = log c/log abc

Consider

1/x + 1/y + 1/z

= log a/log abc + log b/log abc + log c/log abc

= (log a + log b + log c)/log abc

Using Additive Property of logarithm

= log abc/log abc

= 1

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

Simplifying we get

xy + yz + zx = xyz

Hence, Proved