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

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Answer 1 · 2018-08-03T13:04:29

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

We will using that $log(a)-log(b)=log(a/b)$ writing

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

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

$(x-1)/(3x)=e^(12+log(2))$

let

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

then we get

$(x-1)/(3x)=c$

Multiplying by $3x$

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

so $x=1/(1-3c)$

but the is negative, so no solutions!

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