How do you solve $2 ln x + 3 ln 2 = 5$ $2lnx+3ln2=5$ ?

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How do you solve $2 ln x + 3 ln 2 = 5$ $2lnx+3ln2=5$ ?

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Answer 1 · 2015-12-15T08:00:19

$x ≅ 4.30716$ $x ~= 4.30716$

Explanation:

Property of Logarithmic expression

$log A + log B = log (A B) (1)$ $log A + log B = Log(AB) " " " " " (1)$
$n log A = {log A}^{n} (2)$ $n log A= log A^n " " " " (2)$

Given :

$2 ln x + 3 ln 2 = 5$ $2ln x + 3ln 2 = 5$

Rewrite as:
Using rule (1)

${ln x}^{2} + {ln 2}^{3} = 5$ $lnx^2 + ln2^3 = 5$

Using rule (1)

$ln (x^{2} \cdot 8) = 5$ $ln(x^2 * 8) = 5$

Raise the expression to exponential form, with the base of $e$ $e$

$e^{ln (8 x^{2}) = e^{5}}$ $e^(ln(8x^2) = e^5$

$8 x^{2} = e^{5}$ $8x^2 = e^5$
$x^{2} = \frac{e^{5}}{8}$ $x^2 = (e^5)/8$

$x = \pm \sqrt{\frac{e^{5}}{8}}$ $x = +-sqrt((e^5)/8)$

$x ≅ 4.30716$ $x ~= 4.30716$

Because the argument of any logarithm always POSITIVE and greater than zero, due to domain restriction.

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How do you solve $2 ln x + 3 ln 2 = 5$ $2lnx+3ln2=5$ ?

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How do you solve $2 ln x + 3 ln 2 = 5$ $2lnx+3ln2=5$ ?