Find x if :
${log}_{7} (2 x^{2} - 1) = 2$

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Solution

Consider the equation
Find x if :
${log}_{7} (2 x^{2} - 1) = 2$
$\Rightarrow 7^{2} = 2 x^{2} - 1$
$\Rightarrow 7 \times 7 = 2 x^{2} - 1$
$\Rightarrow 2 x^{2} - 1 - 49 = 0$
$\Rightarrow 2 x^{2} - 50 = 0$
$\Rightarrow 2 x^{2} = 50$
$\Rightarrow x^{2} = \frac{50}{2}$
$\Rightarrow x^{2} = 25$
$\Rightarrow x - \pm \sqrt{25}$
$\Rightarrow x = 5$ [ neglecting the negative value]

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Similar questions

Find x, if:

(i) $l o g_{3} x = 0$

(ii) $l o g_{x} 2 = - 1$

(iii) $l o g_{9} 243 = x$

(iv) $l o g_{5} (x - 7) = 1$

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(vi) $l o g_{7} (2 x^{2} - 1) = 2$

x - \frac{1}{x} = - 1

x^{2} + \frac{1}{x^{2}}

{log}_{7} (2 x^{2} - 1) = 2

x

x + 1 / x = 3

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x = 2 + \sqrt{3} f i n d x + \frac{1}{x}

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