Solve for x natural log of x+ natural log of x^2=5

$\ln (x) + \ln (x^{2}) = 5$

Simplify the left side.

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Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

Multiply $x$ by $x^{2}$ by adding the exponents.

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Multiply $x$ by $x^{2}$ .

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Raise $x$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\ln (x^{1 + 2}) = 5$

Add $1$ and $2$ .

$\ln (x^{3}) = 5$

The exponent of a factor inside a logarithm can be expanded to the front of the expression using the third law of logarithms. The third law of logarithms states that the logarithm of a power of $x$ is equal to the exponent of that power times the logarithm of $x$ $(\log_{b} (x^{n}) = n \log_{b} (x))$ .

$3 \ln (x) = 5$

Divide each term by $3$ and simplify.

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Divide each term in $3 \ln (x) = 5$ by $3$ .

$\frac{3 \ln (x)}{3} = \frac{5}{3}$

Cancel the common factor of $3$ .

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Cancel the common factor.

$\frac{3 \ln (x)}{3} = \frac{5}{3}$

Divide $\ln (x)$ by $1$ .

$\ln (x) = \frac{5}{3}$

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (x)} = e^{\frac{5}{3}}$

Exponentiation and log are inverse functions.

$x = e^{\frac{5}{3}}$

The result can be shown in multiple forms.

Exact Form:

$x = e^{\frac{5}{3}}$

Decimal Form:

$x = 5.29449005 \dots$

Solve for x natural log of x+ natural log of x^2=5

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