How do you solve for x for $ln 1 - ln e = x$ $ln1 - lne = x$ ?

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Answer 1 · 2015-02-17T16:28:28

The result should be: $x = - 1$ $x=-1$

Consider that from the definition of logarithm:

${log}_{a} b = x \Rightarrow a^{x} = b$ $log_ab=x =>a^x=b$

and that ${log}_{e}$ $log_e$ is $ln$ $ln$ ;

So you get:

$ln 1 = 0$ $ln1=0$ because $e^{0} = 1$ $e^0=1$

$ln e = 1$ $lne=1$ because $e^{1} = e$ $e^1=e$

in your equation.:

$0 - 1 = x$ $0-1=x$

$x = - 1$ $x=-1$

How do you solve for x for ln1 - lne = xln1−lne=xln1 - lne = x?