Question

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

Question :

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

Solve it .

Solution :

$\sf \: \dfrac{1}{2} log_{ \sqrt{3} }( \dfrac{x + 1}{x + 5} ) + log_{9} {(x + 5)}^{2} = 1 \\ \\ \implies \sf \: \dfrac{1}{2} . \dfrac{ log( \dfrac{x + 1}{x + 5} ) }{ log( \sqrt{3} ) } + \dfrac{ log( {x + 5)}^{2} }{ log(9) } \: \: \{identity \: : log_{a}(b) = \dfrac{ log(b) }{ log(a) } \} \\ \\ \implies \sf \: \dfrac{1}{2} . \dfrac{ log( \dfrac{x + 1}{x + 5} ) }{ \dfrac{1}{2} log(3) } + \dfrac{2 log(x + 5) }{2 log(3) } = 1 \\ \\ \implies \sf \: \dfrac{ log( \dfrac{x + 1}{x + 5} ) }{ log(3) } + \dfrac{ log( x + 5) }{ log(3) } = 1 \\ \\ \implies \sf \: log_{3}( \dfrac{x + 1}{x + 5} ) + log_{3}(x + 5) = 1 \: \: \{identity \: : \dfrac{ log(b) }{ log(a) } = log_{a}(b) \} \\ \\ \implies \sf \: log_{3}( \dfrac{x + 1}{x + 5}.(x + 5) ) = 1 \: \{identity \: : log_{a}(m) + log_{a}(n) = log_{a}(mn) \} \\ \\ \implies\sf\:log_3(x+1)=1\:\\ \\ \implies\sf\:x+1=3\:\{Identity\::log_a(b)=x\implies\:a^x=b\} \\ \\ \implies\sf\:x=3-1\\ \\ \implies\sf\:x=2$

Therefore, the value of x is 2.

Answer 2

$\huge \mathfrak{Answer:-}$

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

$\large\underline\mathrm{Solution}$

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

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

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

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

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

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

$\implies$ x + 1 = 3

$\implies$ x = 3 -1

$\implies$ x = 2

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

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

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