How do you solve $log (x - 1) - log 2 = 12 + log 3 x$ $log(x-1) - log2 = 12 + log3x$ ?

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Answer 1 · 2018-08-03T13:04:29

$x = \frac{1}{1 - 3 c}$ $x=1/(1-3c)$ where $c = e^{12 + log (2)}$ $c=e^(12+log(2))$

We will using that $log (a) - log (b) = log (\frac{a}{b})$ $log(a)-log(b)=log(a/b)$ writing

$log (x - 1) - log (3 x) = 12 + log (2)$ $log(x-1)-log(3x)=12+log(2)$ so

$log (\frac{x - 1}{3 x}) = 12 + log (2)$ $log((x-1)/(3x))=12+log(2)$ and

$\frac{x - 1}{3 x} = e^{12 + log (2)}$ $(x-1)/(3x)=e^(12+log(2))$

let

$c = e^{12 + log (2)}$ $c=e^(12+log(2))$

then we get

$\frac{x - 1}{3 x} = c$ $(x-1)/(3x)=c$

Multiplying by $3 x$ $3x$

$x - 1 = 3 x c$ $x-1=3xc$ so $x - 3 x c = 1$ $x-3xc=1$

so $x = \frac{1}{1 - 3 c}$ $x=1/(1-3c)$

but the is negative, so no solutions!

How do you solve log(x-1) - log2 = 12 + log3x log(x−1)−log2=12+log3xlog(x-1) - log2 = 12 + log3x ?