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If $(0.2)^{x}=2$ and $\log _{10} 2=0.3010$, then what is the value of x to the nearest tenth?
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The correct answer is:
$-0.4$
$(0.2)^{x}=2$
Taking log on both sides,
$x \log _{10} \frac{2}{10}=\log _{10} 2$
$x\left[\log _{10} 2-\log _{10} 10\right]=\log _{10} 2$
$x[0.3010-1]=0.3010$
$x=-\frac{0.3010}{0.6990}=-0.43$
Taking log on both sides,
$x \log _{10} \frac{2}{10}=\log _{10} 2$
$x\left[\log _{10} 2-\log _{10} 10\right]=\log _{10} 2$
$x[0.3010-1]=0.3010$
$x=-\frac{0.3010}{0.6990}=-0.43$
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