Calculus Examples

$- x^{- 2} \ln (x) + \frac{x^{- 1}}{x} = 0$

Step 1

Simplify each term.

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Step 1.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- \frac{1}{x^{2}} \ln (x) + \frac{x^{- 1}}{x} = 0$

Step 1.2

Combine $\ln (x)$ and $\frac{1}{x^{2}}$ .

$- \frac{\ln (x)}{x^{2}} + \frac{x^{- 1}}{x} = 0$

Step 1.3

Move $x^{- 1}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- \frac{\ln (x)}{x^{2}} + \frac{1}{x \cdot x} = 0$

Step 1.4

Multiply $x$ by $x$ .

$- \frac{\ln (x)}{x^{2}} + \frac{1}{x^{2}} = 0$

Step 2

Subtract $\frac{1}{x^{2}}$ from both sides of the equation.

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

Step 3

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$- \ln (x) = - 1$

Step 4

Divide each term in $- \ln (x) = - 1$ by $- 1$ and simplify.

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Step 4.1

Divide each term in $- \ln (x) = - 1$ by $- 1$ .

$\frac{- \ln (x)}{- 1} = \frac{- 1}{- 1}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Dividing two negative values results in a positive value.

$\frac{\ln (x)}{1} = \frac{- 1}{- 1}$

Step 4.2.2

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

$\ln (x) = \frac{- 1}{- 1}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $- 1$ by $- 1$ .

$\ln (x) = 1$

Step 5

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

$e^{\ln (x)} = e^{1}$

Step 6

Rewrite $\ln (x) = 1$ 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 $b^{y} = x$ .

$e^{1} = x$

Step 7

Rewrite the equation as $x = e^{1}$ .

$x = e$

Step 8

The result can be shown in multiple forms.

Exact Form:

$x = e$

Decimal Form:

$x = 2.71828182 \dots$