Question

100 points ...


please help

Answer 1

Helo...

refer attachment
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thanking you..

#Vaish...


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Answer 2

hiii......friend

the answer is here,

$= > \: \frac{ log_{3}(135) }{ log_{15}(3) } - \frac{ log_{3}(5) }{ log_{405}(3) }$

We know that,

$= > \: \frac{1}{ log_{a}(b) } = log_{b}(a)$

$= > \: log_{3}(135) \times log_{3}(15) - log_{3}(5) \times log_{3}(405)$

$= > \: log_{3}( {3}^{3} \times 5 ) \times log_{3}(5 \times 3) - log_{3}(5) \times log_{3}( {3}^{4} \times 5)$

$= > \:( 3 log_{3}(3) + log_{3}(5)) \times ( log_{3}(5) + log_{3}(3) ) - log_{3}(5) \times( 4 log_{3}(3) + log_{3}(5) )$

$= > \: (3 + log_{3}(5) )( log_{3}(5) + 1) - log_{3}(5) (4 + log_{3}(5) )$

$let \: \: log_{3}(5) \: be \: \: x$

$= > \: (3 + x)(x + 1) - x(x + 4)$

$= > \: {x}^{2} + 4x + 3 - {x}^{2} - 4x$

$= > \: 3$

So, The answer is 3.

:-)Hope it helps u.