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Logarithms

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Question : 40 of 85

Marks: +1, -0

If

\log_{30} 3 = a, \log_{30} 5 = b,

\log_{30} 8 =

$3(1-a-b)$
$a-b+1$
$1-a-b$
$1(a-b+1)$ .

Solution:

Concept:

Express

\log_{30}8

using

\log_{30}2

, then find

\log_{30}2

from the product of factors of 30.

Explanation:

Given

\log_{30}3 = a

and

\log_{30}5 = b

.

We have

30 = 2 \times 3 \times 5

.

Take

\log_{30}

on both sides:

\log_{30}30 = \log_{30}2 + \log_{30}3 + \log_{30}5

.

Since

\log_{30}30 = 1

, we get

1 = \log_{30}2 + a + b

.

Thus,

\log_{30}2 = 1 - a - b

.

Now,

\log_{30}8 = \log_{30}(2^3) = 3\log_{30}2 = 3(1 - a - b)

.

Answer:

3(1 - a - b)

, which corresponds to option A.

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