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Logarithms

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Question : 54 of 85

Marks: +1, -0

If

a=\log_{24}12, b=\log_{36}24,

c=\log_{48}36,

1+abc

is equal to :

$2bc$
$2ab$
$2ac$
none

Solution:

Concept:

Use change-of-base to simplify the product

abc

.

Explanation:

Given

a = \log_{24}12

,

b = \log_{36}24

,

c = \log_{48}36

.

Using change of base (any common base, e.g., natural log):

a = \frac{\ln 12}{\ln 24}

,

b = \frac{\ln 24}{\ln 36}

,

c = \frac{\ln 36}{\ln 48}

.

Then

abc = \frac{\ln 12}{\ln 24} \cdot \frac{\ln 24}{\ln 36} \cdot \frac{\ln 36}{\ln 48} = \frac{\ln 12}{\ln 48} = \log_{48}12

.

Now

1+abc = 1 + \log_{48}12 = \log_{48}48 + \log_{48}12 = \log_{48}(48 \times 12) = \log_{48}576

.

Note that

b = \log_{36}24

, but compute

bc = \left(\frac{\ln 24}{\ln 36}\right)\left(\frac{\ln 36}{\ln 48}\right) = \frac{\ln 24}{\ln 48} = \log_{48}24

.

Thus

2bc = 2\log_{48}24 = \log_{48}(24^2) = \log_{48}576

.

Therefore

1+abc = 2bc

.

Answer:

1+abc = 2bc

, which corresponds to option A.

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