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Logarithms

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Question : 22 of 85

Marks: +1, -0

If

x = \log_{2a} a, y = \log_{3a} 2a

z = \log_{4a} 3a

, then the value of

y z(2-x)

is:

1
$-1$
2
-2

Solution:

Concept:

Use change of base to express all logs in a common base, then simplify the product.

Explanation:

Set all logs to base 10 (or natural).

x = \log_{2a} a = \frac{\log a}{\log(2a)}

,

y = \log_{3a} 2a = \frac{\log(2a)}{\log(3a)}

,

z = \log_{4a} 3a = \frac{\log(3a)}{\log(4a)}

.

Then

yz = \frac{\log(2a)}{\log(3a)} \cdot \frac{\log(3a)}{\log(4a)} = \frac{\log(2a)}{\log(4a)}

.

Now

2 - x = 2 - \frac{\log a}{\log(2a)} = \frac{2\log(2a) - \log a}{\log(2a)}

. But

2\log(2a) = \log((2a)^2) = \log(4a^2)

, so

2\log(2a) - \log a = \log(4a^2) - \log a = \log\left(\frac{4a^2}{a}\right) = \log(4a)

.

Thus

2-x = \frac{\log(4a)}{\log(2a)}

.

Finally,

yz(2-x) = \frac{\log(2a)}{\log(4a)} \cdot \frac{\log(4a)}{\log(2a)} = 1

.

Answer:

1

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