Question
Given $2 \log_{10}x + 1 = \log_{10}250,$ find $:(i) x,(ii) \log_{10}2x$
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Answer
$(i)$ Consider the given equation:
$2 \log _{10} x+1=\log _{10} 250$
$\Rightarrow \log _{10} x^2+1=\log _{10} 250 \quad\left[\log _a m^n=n \log _a m\right]$
$\Rightarrow \log _{10} x 2+\log _{10} 10=\log _{10} 250\left[\because \log _{10} 10=1\right]$
$\Rightarrow \log _{10}\left(x^2 \times 10\right)=\log _{10} 250 \quad\left[\log _a m+\log _a n=\log _a m n\right]$
$\Rightarrow x^2 \times 10=250$
$\Rightarrow x^2=25$
$\Rightarrow x=\sqrt{25}$
$\Rightarrow x=5$
$(ii) x=5 ($proved above in $(i))$
$\log _{10} 2 x=\log _{10} 2(5)$
$=\log _{10} 10$
$=1\left[\because \log _{10} 10=1\right]$
$2 \log _{10} x+1=\log _{10} 250$
$\Rightarrow \log _{10} x^2+1=\log _{10} 250 \quad\left[\log _a m^n=n \log _a m\right]$
$\Rightarrow \log _{10} x 2+\log _{10} 10=\log _{10} 250\left[\because \log _{10} 10=1\right]$
$\Rightarrow \log _{10}\left(x^2 \times 10\right)=\log _{10} 250 \quad\left[\log _a m+\log _a n=\log _a m n\right]$
$\Rightarrow x^2 \times 10=250$
$\Rightarrow x^2=25$
$\Rightarrow x=\sqrt{25}$
$\Rightarrow x=5$
$(ii) x=5 ($proved above in $(i))$
$\log _{10} 2 x=\log _{10} 2(5)$
$=\log _{10} 10$
$=1\left[\because \log _{10} 10=1\right]$
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