How do you solve $ln x - ln 5 = 0$ $lnx-ln5=0$ ?

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Answer 1 · 2017-01-04T04:07:24

$x = 5$ $x=5$

Using the property that $a^{{log}_{a} (x)} = x$ $a^(log_a(x))=x$ and $ln$ $ln$ is the base- $e$ $e$ logarithm:

$ln (x) - ln (5) = 0$ $ln(x)-ln(5) = 0$

$\Rightarrow ln (x) = ln (5)$ $=> ln(x) = ln(5)$

$\Rightarrow e^{ln (x)} = e^{ln (5)}$ $=> e^ln(x) = e^ln(5)$

$:. x = 5$

Answer 2 · 2017-01-04T04:59:40

$lnx-ln5=0$

Add $ln5$ to each side:
$lnx=ln5$

$x=5$

How do you solve lnx-ln5=0lnx−ln5=0lnx-ln5=0?