Question

solve:
$log_{3 - 4 {x}^{2} }(9 - 16 {x}^{4} ) = 2 + \frac{1}{ log_{2}(3 - 4 {x}^{2} ) }$
​

Answer 1

Step-by-step explanation:

log3(x+5x+1)+log9(x+5)2=1

Solve it .

Solution :

\sf\frac{1}{2} log_{ \sqrt{3} }( \frac{x + 1}{x + 5} ) + log_{9}( {x + 5)}^{2} = 121log3(x+5x+1)+log9(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) } ]⟹21.log(3)log(x+5x+1)+log(9)log(x+5)2=1[identity:loga(b)=log(a)log(b)]

\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⟹21.21log(3)log(x+5x+1)+2log(3)2log(x+5)=1

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

\implies \sf \: log_{3}( \frac{x + 1}{x + 5} ) + log_{3}(x + 5) = 1 \: \: [identity : \frac{ log(b) }{ log(a) } = log_{a}(b) ]⟹log3(x+5x+1)+log3(x+5)=1[identity:log(a)log(b)=loga(b)]

\implies \sf \: log_{3}( \frac{x + 1}{x + 5} .( x + 5)) = 1 \: \: [identity : log_{a}(m) + log_{a}(n) = log_{a}(mn) ]⟹log3(x+5x+1.(x+5))=1[identity:loga(m)+loga(n)=loga(mn)]

\implies \sf \: log_{3}(x + 1) = 1⟹log3(x+1)=1

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

\implies \sf \: x = 3 - 1⟹x=3−1

\implies\sf{x=2}⟹x=2

Therefore, the value of x is 2.