How do you solve $2 ln x + {ln x}^{2} = 3$ $2 ln x+ ln x^2=3$ ?

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How do you solve $2 ln x + {ln x}^{2} = 3$ $2 ln x+ ln x^2=3$ ?

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Answer 1 · 2015-12-04T11:19:54

$x = e^{\frac{3}{4}}$ $x = e^(3/4)$

Explanation:

First of all, you need to "unite" the $ln$ $ln$ expressions into one.

This can be done with the logarithmic rules:

${log}_{a} (n) + {log}_{a} (m) = {log}_{a} (n \cdot m)$ $log_a (n) + log_a (m) = log_a (n * m )$

$r \cdot {log}_{a} (n) = {log}_{a} (n^{r})$ $r * log_a(n) = log_a(n^r)$

So, you can transform your equation as follows:

$\times 2 ln x + {ln x}^{2} = 3$ $color(white)(xx)2 ln x + ln x^2 = 3$

$\Leftrightarrow 2 ln x + 2 ln x = 3$ $<=> 2 ln x + 2 ln x = 3$

$\Leftrightarrow 4 ln x = 3$ $<=> 4 ln x = 3$

... divide both sides by $4$ $4$ ...

$\Leftrightarrow ln x = \frac{3}{4}$ $<=> ln x = 3/4$

Now, the inverse function for $ln x$ $ln x$ is $e^{x}$ $e^x$ which means that both $ln (e^{x}) = x$ $ln(e^x) = x$ and $e^{ln x} = x$ $e^(ln x)= x$ hold.

This means that you can apply $e^{x}$ $e^x$ to both sides of the equation to "get rid" of the logarithm:

$\Leftrightarrow e^{ln (x)} = e^{\frac{3}{4}}$ $<=> e^ln(x) = e^(3/4)$

$\Leftrightarrow x = e^{\frac{3}{4}}$ $<=> x = e^(3/4)$

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