How do you solve $Log (x-1) + Log 2 = Log (3x)$ ?

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Answer 1 · 2018-04-10T03:20:51

There is no solution for $x$ .

By the laws of logarithms, $loga + logb = log(a*b)$ .

Using this law:
$log(x-1)+log2 = log(3x)$
$log(2(x-1)) = log(3x)$

Using the distributive property ( $a*(b+c) = ab + ac$ ):
$log(2x -2) = log(3x)$

$10^(log(2x-2)) = 10^log(3x)$
$2x-2 = 3x$
$3x = 2x -2$
Subtracting $2x$ from both sides:
$x = -2$

Since you can not apply $log$ on a negative number, there is no solution for this equation.

How do you solve Log (x-1) + Log 2 = Log (3x)Log (x-1) + Log 2 = Log (3x)?