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Logarithms

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Question : 47 of 85

Marks: +1, -0

If

\log(x+y)=\log\left(\;\frac{3x-3y}{2}\right)

\log x-\log y=

$\log 2$
$\log 3$
$\log 5$
$\log 6$ .

Solution:

Concept:

If

\log A = \log B

, then

A = B

(for same base).

Explanation:

Given

\log(x+y) = \log\left(\frac{3x-3y}{2}\right)

.

Equate the arguments:

x+y = \frac{3x-3y}{2}

.

Multiply both sides by 2:

2(x+y) = 3x-3y

.

Simplify:

2x+2y = 3x-3y

.

Bring like terms:

2y+3y = 3x-2x

→

5y = x

.

Thus

x = 5y

.

Then

\log x - \log y = \log\left(\frac{x}{y}\right) = \log\left(\frac{5y}{y}\right) = \log 5

.

Answer:

\log 5

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