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Logarithms

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Question : 82 of 85

Marks: +1, -0

If

\log_{7} \log_{5}(\sqrt{x+5}+\sqrt{x})=0,

x

is

0
1
$\frac{1}{4}$
4

Solution:

Concept:

Using the property

\log_a b = 0 \Rightarrow b = 1

and

\log_a b = 1 \Rightarrow b = a

.

Explanation:

Given:

\log_{7} \log_{5}(\sqrt{x+5} + \sqrt{x}) = 0

.

Then

\log_{5}(\sqrt{x+5} + \sqrt{x}) = 7^0 = 1

.

So

\sqrt{x+5} + \sqrt{x} = 5^1 = 5

.

Let

\sqrt{x+5} = a

and

\sqrt{x} = b

. Then

a + b = 5

, and

a^2 - b^2 = (x+5) - x = 5

.

Since

a^2 - b^2 = (a-b)(a+b) = 5

, we have

(a-b) \cdot 5 = 5 \Rightarrow a - b = 1

.

Solve

a + b = 5

and

a - b = 1

: adding gives

2a = 6 \Rightarrow a = 3

, then

b = 2

.

Thus

\sqrt{x} = 2 \Rightarrow x = 4

.

Answer:

x = 4

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