If x = loga (bc), y = log4 (ca) and z = logc(ab) when which of the following is equal to 1?a)x + y + zb)(1 + x )-1+ (1+ y)-1+ (1+ z)-1c)xyzd)None of theseCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

To understand why option B is the correct answer, let's first analyze the given equations:

x = logₐ(bc)
y = logₑ(ca)
z = logc(ab)

We can rewrite these equations using exponentiation:

aˣ = bc
bʸ = ca
cᶻ = ab

Now, let's try to simplify the expressions:

xyz = (aˣ)(bʸ)(cᶻ)

To simplify this, we need to find a common base. Since we have a, b, and c in the equations, we can rewrite them as:

xyz = (aᶻ)(bˣ)(cʸ)

Now we can substitute the values of x, y, and z:

xyz = (a(logₐ(bc)))((b(logₑ(ca))))((c(logc(ab))))

Using the properties of logarithms, we can simplify this to:

xyz = (a(logₐ(b)) + a(logₐ(c)))(b(logₑ(c)) + b(logₑ(a)))(c(logc(a)) + c(logc(b)))

Further simplifying:

xyz = (logₐ(b) + logₐ(c))(logₑ(c) + logₑ(a))(logc(a) + logc(b))

Now, let's analyze option B:

(1/x) = (1/logₐ(bc))
(1/y) = (1/logₑ(ca))
(1/z) = (1/logc(ab))

We can substitute the values of x, y, and z:

(1/x) = (1/(logₐ(bc))) = (1/(logₐ(b) + logₐ(c)))
(1/y) = (1/(logₑ(ca))) = (1/(logₑ(c) + logₑ(a)))
(1/z) = (1/(logc(ab))) = (1/(logc(a) + logc(b)))

Now, let's analyze the expression in option B:

(1/x)(1/y)(1/z) = ((1/(logₐ(b) + logₐ(c))))((1/(logₑ(c) + logₑ(a))))((1/(logc(a) + logc(b))))

Using the properties of logarithms, we can simplify this to:

(1/x)(1/y)(1/z) = (1/(logₐ(b) + logₐ(c)))(1/(logₑ(c) + logₑ(a)))(1/(logc(a) + logc(b)))

Comparing this expression with the simplified expression of xyz, we can see that they are equal. Therefore, option B is the correct answer.

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