Solve for x log base 6 of x+ log base 6 of x^2+ natural log of e=2

$\log_{6} (x) + \log_{6} (x^{2}) + \ln (e) = 2$

Step 1

Simplify $\log_{6} (x) + \log_{6} (x^{2}) + \ln (e)$ .

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

$\log_{6} (x \cdot x^{2}) + \ln (e) = 2$

Simplify each term.

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

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

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

$\log_{6} (x^{1} x^{2}) + \ln (e) = 2$

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

$\log_{6} (x^{1 + 2}) + \ln (e) = 2$

Add $1$ and $2$ .

$\log_{6} (x^{3}) + \ln (e) = 2$

The natural logarithm of $e$ is $1$ .

$\log_{6} (x^{3}) + 1 = 2$

Step 2

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x \approx 1.81712059$

Step 3