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Logarithms

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Question : 72 of 85

Marks: +1, -0

\log(a-9)+\log a=1,

the value of 'a' is

$0$
$10$
$-1$
None of these

Solution:

Concept:

Use the logarithm product rule and convert to exponential form.

Explanation:

Given:

\log(a-9) + \log a = 1

.

Using

\log x + \log y = \log(xy)

:

\log[(a-9)a] = 1

.

Convert to exponential (base 10):

(a-9)a = 10^1 = 10

.

So

a^2 - 9a - 10 = 0

.

Factor:

(a-10)(a+1) = 0

⇒

a = 10

or

a = -1

.

Arguments must be positive:

a-9 > 0

and

a > 0

⇒

a > 9

.

Thus only

a = 10

is valid.

Answer:

a = 10

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