How do you solve ${ln (x - 2)}^{2} = 6$ $ln(x-2)^2=6$ ?

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How do you solve ${ln (x - 2)}^{2} = 6$ $ln(x-2)^2=6$ ?

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Answer 1 · 2016-08-15T06:30:03

We will solve for $x$ $x$ using the very basic properties of natural logarithms. I'll illustrate it below.

Explanation:

We have, ${ln (x - 2)}^{2} = 6$ $Ln (x - 2)^2 = 6$

Using the property, ${ln m}^{n} = n ln m$ $Ln m^n = nLn m$ ,

We have, ${ln (x - 2)}^{2} = 6$ $Ln (x - 2)^2 = 6$

$\Rightarrow 2 ln (x - 2) = 6$ $implies 2Ln (x - 2) = 6$
$\Rightarrow ln (x - 2) = 3$ $implies Ln (x - 2) = 3$
$\Rightarrow x - 2 = e^{3}$ $implies x-2 = e^3$

In the last step, we used the concept of inverse of the natural logarithms, $ln m = n \Rightarrow m = e^{n}$ $Ln m = n implies m = e^n$

Thus, $x = e^{3} + 2$ $x = e^3 + 2$

Where $e$ $e$ is the base of natural logarithms.

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How do you solve ${ln (x - 2)}^{2} = 6$ $ln(x-2)^2=6$ ?

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How do you solve ln(x−2)2=6ln(x-2)^2=6?

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Explanation:

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How do you solve ${ln (x - 2)}^{2} = 6$ $ln(x-2)^2=6$ ?