Question

:) log
35/
78
= log 7 + log 5 - log 2 - log 3 - log. 13​

Answer 1

Answer:

log 3578 log 7 + log 5 - log to log 3 - 13

Answer 2

Answer:

Given,

$log( \frac{35}{78} )$

We want to prove that

$log( \frac{35}{78} ) = log(7) + log(5) - log(2) - log(3) - log(13)$

We know,

$log( \frac{x}{y} ) = log(x) - log(y)$

So,

$log( \frac{35}{78} ) = log(35) - log(78) \\ = log(5 \times 7) - log(2 \times 3 \times 13)$

We know,

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

So,

$log(5 \times 7) = log(5) + log(7) \\ log(2 \times 3 \times 13) = log(2) + log(3) + log(13)$

So,

$log( \frac{35}{78} ) \\ = log(3) + log(5) - ( log(2) + log(3) + log(13)) \\ = log(3) + log(5) - log(2) - log(3) - log(13)$

Therefore,it is proved that

$log( \frac{35}{78} ) = log(7) + log(5) - log(2) - log(3) - log(13)$

This is a problem of Logarithm formulas,

Some important Logarithm formulas are

$log_{x}(1) = 0 \\ log_{x}(0) = 1 \\ log_{x}(y) = \frac{ log(x) }{ log(y) } \\ log( {x}^{y} ) = y log(x) \\ log(x) + log(y) = log(xy) \\ log(x) - log(y) = log( \frac{x}{y} ) \\ log_{x}(x) = 1$

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