Question

Log (base 7) 2 + log (base 49) x = log (base 1/7) sqrt 3

Answer 1

Answer:

$(log \sqrt5 to \: the \: base \: of \: 7\: (log \: 49 \: to \: the \: base \: of \: 5)$

Answer 2

Answer:

Required value of x is $\frac{ \sqrt{2} }{6}$

Step-by-step explanation:

Given,

$log_{7}(2) + log_{49}(x) = log_{ \frac{1}{7} }( \sqrt{3} ) \\ \frac{ log(2) }{ log(7) } + \frac{ log(x) }{ log(49) } = \frac{ log( \sqrt{3} ) }{ log( \frac{1}{7} ) } \\ \frac{ log(2) }{ log(7) } + \frac{ log(x) }{ log( {7}^{2} ) } = \frac{ log( \sqrt{3} ) }{ - log(7) } \\ \frac{ log(2) }{ log(7) } + \frac{ log(x) }{ 2log(7) } = - \frac{ log( \sqrt{3} ) }{ log(7) } \\ \frac{ 2log(2) + log(x) }{2 log(7) } = - \frac{ log( \sqrt{3} ) }{ log(7) } \\ 2log(2) + log(x) = - 2 log( \sqrt{3} ) \\ log(x) = - 2 log( \sqrt{3} ) - 2 log(2) \\ log(x) = - 2(log( \sqrt{3} ) + log(2)) \\ log(x) = - 2 log(2 \times \sqrt{3} ) \\ log(x) = - log( \sqrt{2 \sqrt{3} } ) \\ log(x) = log( \frac{1}{3 \sqrt{2} } ) \\ x = \frac{1}{3 \sqrt{2} }$

Therefore,

$x = \frac{1}{3 \sqrt{2} } \\ x = \frac{ \sqrt{2} }{3 \sqrt{2} \times \sqrt{2} } \\ x = \frac{ \sqrt{2} }{3 \times 2} \\ x = \frac{ \sqrt{2} }{6}$

Here applied formulas are

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

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