Question

solve log_(3-4x^(2))(9-16x^(4))=2+(1)/(log _(2)(3-4x^(2)))​

Answer 1

$\textbf{Given:}$

$\log_{3-4\,x^2}(9-16\,x^4)=2+\dfrac{1}{\log_2(3-4\,x^2)}$

$\textbf{To find:}$

$\text{The value of x}$

$\textbf{Solution:}$

$\text{Consider,}$

$\log_{3-4\,x^2}(9-16\,x^4)=2+\dfrac{1}{\log_2(3-4\,x^2)}$

$\text{Using base changing formula,}$

$\log_{3-4\,x^2}(9-16\,x^4)=2+\log_{3-4\,x^2}2$

$\log_{3-4\,x^2}(9-16\,x^4)-\log_{3-4\,x^2}2=2$

$\text{Using quotient rule of logarithm,}$

$\log_{3-4\,x^2}(\frac{9-16\,x^4}{2})=2$

$\implies(3-4\,x^2)^2=\dfrac{9-16\,x^4}{2}$

$\implies\;9+16\,x^4-24\,x^2=\dfrac{9-16\,x^4}{2}$

$\implies\;2(9+16\,x^4-24\,x^2)=9-16\,x^4$

$\implies\;18+32\,x^4-48\,x^2=9-16\,x^4$

$\implies\;48\,x^4-48\,x^2+9=0$

$\implies\;16\,x^4-16\,x^2+3=0$

$\implies\;16\,x^4-12\,x^2-4\,x^2+3=0$

$\implies\;4x^2(4\,x^2-3)-1(4\,x^2-3)=0$

$\implies\;(4x^2-1)(4\,x^2-3)=0$

$\implies\;x^2=\dfrac{1}{4},\dfrac{3}{4}$

$\text{But x^2=\dfrac{3}{4} does not satisfy the equation}$

$x^2=\dfrac{1}{4}$

$\implies\,x=\pm\dfrac{1}{2}$

$\textbf{Answer:}$

$\textbf{The values of x are \bf\pm\dfrac{1}{2} }$

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Answer 2

Step-by-step explanation:

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