Question
If $2 \log x + 1 = \log$ $360$, find: $\log (3 x^2 - 8)$
✓
Answer
$\log \left(3 x^2-8\right) $
$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 \left(3 x ^2-8\right) $
$=\log \left\{3(6)^2-8\right\} $
$=\log (108-8) $
$=\log 100 $
$=\log 10^2 $
$=2 \log 10 $
$=2 \times 1$
$= 2.$
$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 \left(3 x ^2-8\right) $
$=\log \left\{3(6)^2-8\right\} $
$=\log (108-8) $
$=\log 100 $
$=\log 10^2 $
$=2 \log 10 $
$=2 \times 1$
$= 2.$
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