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

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

Simplify the left side.

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

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

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

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

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

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

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

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

Add $1$ and $2$ .

$\ln (x^{3}) = 2$

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

$e^{\ln (x^{3})} = e^{2}$

Rewrite $\ln (x^{3}) = 2$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{2} = x^{3}$

Solve for $x$ .

Rewrite the equation as $x^{3} = e^{2}$ .

$x^{3} = e^{2}$

Take the cube root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[3]{e^{2}}$

The result can be shown in multiple forms.

Exact Form:

$x = \sqrt[3]{e^{2}}$

Decimal Form:

$x = 1.94773404 \dots$

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

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