Question
If $2 \log x + 1 =\log 360,$ find $: \log(2 x -2)$
✓
Answer
$\log (2 x-2) $
$2 \log x+1=\log 360 $
$\Rightarrow \log x^2+\log 10=\log 360 $
$\Rightarrow \log \left(10 x^2\right)=\log 360 $
$\Rightarrow 10 x^2=360 $
$\Rightarrow x^2=\frac{360}{10}=36 $
$\Rightarrow x=\sqrt{36}= \pm 6$
As negative value is rejected,
$\therefore x=6 $
$\therefore \log (2 x-2) $
$=\log (2 \cdot 6-2) $
$=\log 10 $
$=1 .$
$2 \log x+1=\log 360 $
$\Rightarrow \log x^2+\log 10=\log 360 $
$\Rightarrow \log \left(10 x^2\right)=\log 360 $
$\Rightarrow 10 x^2=360 $
$\Rightarrow x^2=\frac{360}{10}=36 $
$\Rightarrow x=\sqrt{36}= \pm 6$
As negative value is rejected,
$\therefore x=6 $
$\therefore \log (2 x-2) $
$=\log (2 \cdot 6-2) $
$=\log 10 $
$=1 .$
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