If x satisfies the inequality log _{25} x^2 + | vedclass

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Question

(B) The domain of the expression is $x > 0$. Given inequality: $\log _{25} x^2 + (\log _5 x)^2 < 2$. Using the property $\log _{a^n} b^m = \frac{m}{n}

Vedclass · Accepted Answer

$(\frac{1}{25}, 5)$

Vedclass · Answer

(B) The domain of the expression is $x > 0$. Given inequality: $\log _{25} x^2 + (\log _5 x)^2 Using the property $\log _{a^n} b^m = \frac{m}{n} \log _a b$, we have $\log _{25} x^2 = \log _{5^2} x^2 = \frac{2}{2} \log _5 x = \log _5 x$. Substituting this into the inequality: $\log _5 x + (\log _5 x)^2 Let $y = \log _5 x$. Then $y^2 + y - 2 Factoring the quadratic: $(y + 2)(y - 1) This implies $-2 Substituting back $y = \log _5 x$: $-2 Converting to exponential form: $5^{-2} Thus, $\frac{1}{25} < x < 5$.

If $x$ satisfies the inequality $\log _{25} x^2 + (\log _5 x)^2 < 2$, then $x$ belongs to

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