Solve the equation ${log}_{4} (2 x^{2} + x + 1) - {log}_{4} (2 x - 1) = 1$

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Solution

According to question,

$l o g_{4} (2 x^{2} + x + 1) - l o g_{4} (2 x - 1) = 1$ ............. As , $(l o g a - l o g b = l o g \frac{a}{b})$

$\Rightarrow$ $l o g_{4} (\frac{2 x^{2} + x + 1}{2 x - 1}) = 1$

$\Rightarrow$ $\frac{2 x^{2} + x + 1}{2 x - 1} = 4^{1}$

$\Rightarrow$ $2 x^{2} + x + 1 = 8 x - 4$

$\Rightarrow$ $2 x^{2} - 7 x + 5 = 0$

$\Rightarrow$ $x = \frac{- 1}{2}, 4$

$\Rightarrow$ $x \neq \frac{- 1}{2}$ , as it is not in domain

Hence, $x = 4$

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