Come risolvi $log (3 x + 1) = 2$ $log (3x + 1) = 2$ ?

$x = 33$ $x=33$

Per definizione se $log a = b$ $loga=b$ , noi abbiamo $10^{b} = a$ $10^b=a$ quindi

$log (3 x + 1) = 2$ $log(3x+1)=2$

$\Leftrightarrow (3 x + 1) = 10^{2}$ $hArr(3x+1)=10^2$ or

$3 x + 1 = 100$ $3x+1=100$ or

$3 x = 100 - 1$ $3x=100-1$ or

$3 x = 99$ $3x=99$ or

$x = \frac{99}{3} = 33$ $x=99/3=33$

Come risolvi log (3x + 1) = 2 log(3x+1)=2log (3x + 1) = 2 ?