Question

2 log 5 + 2log 16 - log 3 as log N. slove this ​

Answer 1

SOLUTION

TO EXPRESS

2 log 5 + 2log 16 - log 3 as log N

FORMULA TO BE IMPLEMENTED

1. log a + log b = log ab

$\displaystyle\sf{2. \: \log a - \log b = \log \frac{a}{b} }$

$\sf{ \: 3. \: \: \log {a}^{n} = n \log a}$

EVALUATION

$\displaystyle\sf{2\log 5 + 2\log 16 - \log 3 }$

$\displaystyle\sf{ = \log {5}^{2} + \log {16}^{2} - \log 3 }$

$\displaystyle\sf{ = \log 25 + \log 256 - \log 3 }$

$\displaystyle\sf{ = \log 6400 - \log 3 }$

$\displaystyle\sf{ = \log \frac{6400}{3}}$

Which is of the form log N

Where

$\displaystyle\sf{N = \frac{6400}{3}}$

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Answer 2

HELLO DEAR,

GIVEN:- 2 ㏒ 5 + 2㏒ 16 - ㏒ 3 as ㏒ N.

SOLUTION:- Law of Logarithm with base a.

i) ㏒ (mn) = (㏒ m ) + (㏒ n )

ii) ㏒ (m/n) = (㏒ m ) - (㏒ n )

iii) ㏒ (m^K ) = k . (㏒ m )

iv) ㏒ 1 = 0

v) ㏒ a = 1

So, 2 ㏒ 5 + 2㏒ 16 - ㏒ 3

㏒ 5² + ㏒ 16² - ㏒ 3

⇒ ㏒ 25 + ㏒ 256 - ㏒ 3

⇒ ㏒ (25×256) - ㏒ 3

⇒ ㏒ 6400 - ㏒ 3

⇒ ㏒ (6400/3). is given as ㏒ N

Therefore, N = 6400/3.

THANKS.