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Logarithms

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Question : 29 of 85

Marks: +1, -0

If

x=\log_{a}bc, y=\log_{b}ca, z=\log_{c}ab,

then

$xyz=x+y+z+2$
$xyz=x+y+z+1$
$x+y+z=1$
$xyz=1$

Solution:

Concept:

Use the change-of-base property and algebraic manipulation to find a relation among

x

,

y

, and

z

.

Explanation:

Let

p = \ln a

,

q = \ln b

,

r = \ln c

.

Then

x = \log_a(bc) = \frac{\ln(bc)}{\ln a} = \frac{q+r}{p}

,

y = \frac{r+p}{q}

,

z = \frac{p+q}{r}

.

Compute

xyz = \frac{(q+r)(r+p)(p+q)}{pqr}

.

Expand numerator:

(q+r)(r+p)(p+q) = 2pqr + (p^2q + pq^2 + p^2r + pr^2 + q^2r + qr^2)

.

Thus

xyz = 2 + \frac{p^2q + pq^2 + p^2r + pr^2 + q^2r + qr^2}{pqr}

.

Compute

x+y+z = \frac{q+r}{p} + \frac{r+p}{q} + \frac{p+q}{r} = \frac{p^2q + pq^2 + p^2r + pr^2 + q^2r + qr^2}{pqr}

.

Therefore

xyz = 2 + (x+y+z)

, i.e.,

xyz = x+y+z+2

.

Answer:

x+y+z+2

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