Question

$\frac{1}{2} log_{ \sqrt{3} }( \frac{x + 1}{x + 5} ) + log_{9}( {x + 5)}^{2} = 1$
Solve it .

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Answer 1

Question :

$\sf\frac{1}{2} log_{ \sqrt{3} }( \frac{x + 1}{x + 5} ) + log_{9}( {x + 5)}^{2} = 1$

Solve it .

Solution :

$\sf\frac{1}{2} log_{ \sqrt{3} }( \frac{x + 1}{x + 5} ) + log_{9}( {x + 5)}^{2} = 1$

$\implies \sf \: \frac{1}{2} . \frac{ log( \frac{x + 1}{x + 5} ) }{ log( \sqrt{3}) } + \frac{ log( {x + 5)}^{2} }{ log(9) } = 1 \: \: [identity : \: log_{a}(b) = \frac{ log(b) }{ log(a) } ]$

$\implies \sf \: \frac{1}{2} . \frac{ log( \frac{x + 1}{x + 5} ) }{ \frac{1}{2} log(3) } + \frac{2 log(x + 5) }{2 log(3) } = 1$

$\implies \sf \: \frac{ log( \frac{x + 1}{x + 5} ) }{ log(3) } + \frac{ log(x + 5) }{ log(3) } = 1$

$\implies \sf \: log_{3}( \frac{x + 1}{x + 5} ) + log_{3}(x + 5) = 1 \: \: [identity : \frac{ log(b) }{ log(a) } = log_{a}(b) ]$

$\implies \sf \: log_{3}( \frac{x + 1}{x + 5} .( x + 5)) = 1 \: \: [identity : log_{a}(m) + log_{a}(n) = log_{a}(mn) ]$

$\implies \sf \: log_{3}(x + 1) = 1$

$\implies \sf \: x + 1 = 3 \: \:[identity : log_{a}(b) = x \implies {a}^{x} = b]$

$\implies \sf \: x = 3 - 1$

$\implies\sf{x=2}$

Therefore, the value of x is 2.

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General laws of Logarithm :-

•$\sf{log_a(mn)=log_aM+log_aN}$

•$\sf{log_a(\frac{M}{N})=log_aM-logaN}$

•$\sf{log_aM^n=n\:log_aM}$

•$\sf{log_aM=log_bM\times\:log_ab}$

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Answer 2

$\huge \mathfrak{Answer:-}$

$\implies$ x = 2

$\huge \mathfrak{Solutions:-}$

$\implies$ 1/2 log√3(x + 1 / x + 5) + log9(x + 5)² = 1

$\implies$ 1/2 log(x + 1 / x + 5)/log√3 + log(x + 5)²/log(9) = 1

$\implies$ log(x + 1 / x + 5)/log3 + log(x + 5)²/log(3) = 1

$\implies$ log3(x + 1 / x + 5) + log3(x + 5) = 1

$\implies$ log3(x + 1 / x + 5 × (x + 5))= 1

$\implies$ log 3 × (x + 1) = 1

$\implies$ x + 1 = 3

$\implies$ x = 3 - 1

$\implies$ x = 2

$\large\underline\mathrm{hence,}$

$\large\underline\mathrm{the \: value \: of \: x \: is \: 2.}$

$\large\underline\mathrm{Hope \: it \: helps \: you \: plz \: mark \: me \: brainlist}$