Question
Solve the following$:\log 7 + \log (3x - 2) = \log (x + 3) + 1$
✓
Answer
$\log 7+\log (3 x-2)=\log (x+3)+1$
$\Rightarrow \log 7+\log (3 x-2)-\log (x+3)=1$
$\Rightarrow \log \frac{7 \cdot(3 x-2)}{x+3}=\log 10$
$\Rightarrow \frac{7 \cdot(3 x-2)}{x+3}=10$
$\Rightarrow 21 x-14=10(x+3)$
$\Rightarrow 21 x-10 x=30+14$
$\Rightarrow 11 x=44$
$\Rightarrow x=\frac{44}{ 11}=4 .$
$\Rightarrow \log 7+\log (3 x-2)-\log (x+3)=1$
$\Rightarrow \log \frac{7 \cdot(3 x-2)}{x+3}=\log 10$
$\Rightarrow \frac{7 \cdot(3 x-2)}{x+3}=10$
$\Rightarrow 21 x-14=10(x+3)$
$\Rightarrow 21 x-10 x=30+14$
$\Rightarrow 11 x=44$
$\Rightarrow x=\frac{44}{ 11}=4 .$
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