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$20^{2-3 x^2}=(40 \sqrt{5})^{3 x^2-2}$, then $x$ is equal to
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Verified Answer
The correct answer is:
$\pm \sqrt{\frac{2}{3}}$
We have,
$(20)^{2-3 x^2}=(40 \sqrt{5})^{3 x^2-2}$
On taking log on both sides, we get
$\left(2-3 x^2\right) \log 20=\left(3 x^2-2\right) \log 40 \sqrt{5}$
$\begin{aligned} & \Rightarrow\left(2-3 x^2\right)[2 \log 2+\log 5] \\ & =\left(3 x^2-2\right)\left(3 \log 2+\frac{3}{2} \log 5\right) \\ & \Rightarrow\left(3 x^2-2\right)\left[3 \log 2+\frac{3}{2} \log 5+2 \log 2+\log 5\right]=0 \\ & \Rightarrow\left(3 x^2-2\right)=0 \Rightarrow x^2=\frac{2}{3} \\ & \Rightarrow x= \pm \frac{\sqrt{2}}{3}\end{aligned}$
$(20)^{2-3 x^2}=(40 \sqrt{5})^{3 x^2-2}$
On taking log on both sides, we get
$\left(2-3 x^2\right) \log 20=\left(3 x^2-2\right) \log 40 \sqrt{5}$
$\begin{aligned} & \Rightarrow\left(2-3 x^2\right)[2 \log 2+\log 5] \\ & =\left(3 x^2-2\right)\left(3 \log 2+\frac{3}{2} \log 5\right) \\ & \Rightarrow\left(3 x^2-2\right)\left[3 \log 2+\frac{3}{2} \log 5+2 \log 2+\log 5\right]=0 \\ & \Rightarrow\left(3 x^2-2\right)=0 \Rightarrow x^2=\frac{2}{3} \\ & \Rightarrow x= \pm \frac{\sqrt{2}}{3}\end{aligned}$
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