Calculus Examples

$\frac{d y}{d x} = \frac{12}{{(2 + 3 x)}^{2} e^{2 y}}$ , $y (- 2) = 0$

Step 1

Separate the variables.

Tap for more steps...

Step 1.1

Regroup factors.

$\frac{d y}{d x} = \frac{12}{{(2 + 3 x)}^{2}} \cdot \frac{1}{e^{2 y}}$

Step 1.2

Multiply both sides by $e^{2 y}$ .

$e^{2 y} \frac{d y}{d x} = e^{2 y} (\frac{12}{{(2 + 3 x)}^{2}} \cdot \frac{1}{e^{2 y}})$

Step 1.3

Simplify.

Tap for more steps...

Step 1.3.1

Combine.

$e^{2 y} \frac{d y}{d x} = e^{2 y} \frac{12 \cdot 1}{{(2 + 3 x)}^{2} e^{2 y}}$

Step 1.3.2

Cancel the common factor of $e^{2 y}$ .

Tap for more steps...

Step 1.3.2.1

Factor $e^{2 y}$ out of ${(2 + 3 x)}^{2} e^{2 y}$ .

$e^{2 y} \frac{d y}{d x} = e^{2 y} \frac{12 \cdot 1}{e^{2 y} {(2 + 3 x)}^{2}}$

Step 1.3.2.2

Cancel the common factor.

$e^{2 y} \frac{d y}{d x} = e^{2 y} \frac{12 \cdot 1}{e^{2 y} {(2 + 3 x)}^{2}}$

Step 1.3.2.3

Rewrite the expression.

$e^{2 y} \frac{d y}{d x} = \frac{12 \cdot 1}{{(2 + 3 x)}^{2}}$

Step 1.3.3

Multiply $12$ by $1$ .

$e^{2 y} \frac{d y}{d x} = \frac{12}{{(2 + 3 x)}^{2}}$

Step 1.4

Rewrite the equation.

$e^{2 y} d y = \frac{12}{{(2 + 3 x)}^{2}} d x$

Step 2

Integrate both sides.

Tap for more steps...

Step 2.1

Set up an integral on each side.

$\int e^{2 y} d y = \int \frac{12}{{(2 + 3 x)}^{2}} d x$

Step 2.2

Integrate the left side.

Tap for more steps...

Step 2.2.1

Let $u_{1} = 2 y$ . Then $d u_{1} = 2 d y$ , so $\frac{1}{2} d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

Tap for more steps...

Step 2.2.1.1

Let $u_{1} = 2 y$ . Find $\frac{d u_{1}}{d y}$ .

Tap for more steps...

Step 2.2.1.1.1

Differentiate $2 y$ .

$\frac{d}{d y} [2 y]$

Step 2.2.1.1.2

Since $2$ is constant with respect to $y$ , the derivative of $2 y$ with respect to $y$ is $2 \frac{d}{d y} [y]$ .

$2 \frac{d}{d y} [y]$

Step 2.2.1.1.3

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 1$ .

$2 \cdot 1$

Step 2.2.1.1.4

Multiply $2$ by $1$ .

$2$

Step 2.2.1.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int e^{u_{1}} \frac{1}{2} d u_{1} = \int \frac{12}{{(2 + 3 x)}^{2}} d x$

Step 2.2.2

Combine $e^{u_{1}}$ and $\frac{1}{2}$ .

$\int \frac{e^{u_{1}}}{2} d u_{1} = \int \frac{12}{{(2 + 3 x)}^{2}} d x$

Step 2.2.3

Since $\frac{1}{2}$ is constant with respect to $u_{1}$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} \int e^{u_{1}} d u_{1} = \int \frac{12}{{(2 + 3 x)}^{2}} d x$

Step 2.2.4

The integral of $e^{u_{1}}$ with respect to $u_{1}$ is $e^{u_{1}}$ .

$\frac{1}{2} (e^{u_{1}} + C_{1}) = \int \frac{12}{{(2 + 3 x)}^{2}} d x$

Step 2.2.5

Simplify.

$\frac{1}{2} e^{u_{1}} + C_{1} = \int \frac{12}{{(2 + 3 x)}^{2}} d x$

Step 2.2.6

Replace all occurrences of $u_{1}$ with $2 y$ .

$\frac{1}{2} e^{2 y} + C_{1} = \int \frac{12}{{(2 + 3 x)}^{2}} d x$

Step 2.3

Integrate the right side.

Tap for more steps...

Step 2.3.1

Since $12$ is constant with respect to $x$ , move $12$ out of the integral.

$\frac{1}{2} e^{2 y} + C_{1} = 12 \int \frac{1}{{(2 + 3 x)}^{2}} d x$

Step 2.3.2

Let $u_{2} = 2 + 3 x$ . Then $d u_{2} = 3 d x$ , so $\frac{1}{3} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

Tap for more steps...

Step 2.3.2.1

Let $u_{2} = 2 + 3 x$ . Find $\frac{d u_{2}}{d x}$ .

Tap for more steps...

Step 2.3.2.1.1

Differentiate $2 + 3 x$ .

$\frac{d}{d x} [2 + 3 x]$

Step 2.3.2.1.2

Differentiate.

Tap for more steps...

Step 2.3.2.1.2.1

By the Sum Rule, the derivative of $2 + 3 x$ with respect to $x$ is $\frac{d}{d x} [2] + \frac{d}{d x} [3 x]$ .

$\frac{d}{d x} [2] + \frac{d}{d x} [3 x]$

Step 2.3.2.1.2.2

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [3 x]$

Step 2.3.2.1.3

Evaluate $\frac{d}{d x} [3 x]$ .

Tap for more steps...

Step 2.3.2.1.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$0 + 3 \frac{d}{d x} [x]$

Step 2.3.2.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$0 + 3 \cdot 1$

Step 2.3.2.1.3.3

Multiply $3$ by $1$ .

$0 + 3$

Step 2.3.2.1.4

Add $0$ and $3$ .

$3$

Step 2.3.2.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\frac{1}{2} e^{2 y} + C_{1} = 12 \int \frac{1}{{u_{2}}^{2}} \cdot \frac{1}{3} d u_{2}$

Step 2.3.3

Simplify.

Tap for more steps...

Step 2.3.3.1

Multiply $\frac{1}{{u_{2}}^{2}}$ by $\frac{1}{3}$ .

$\frac{1}{2} e^{2 y} + C_{1} = 12 \int \frac{1}{{u_{2}}^{2} \cdot 3} d u_{2}$

Step 2.3.3.2

Move $3$ to the left of ${u_{2}}^{2}$ .

$\frac{1}{2} e^{2 y} + C_{1} = 12 \int \frac{1}{3 {u_{2}}^{2}} d u_{2}$

Step 2.3.4

Since $\frac{1}{3}$ is constant with respect to $u_{2}$ , move $\frac{1}{3}$ out of the integral.

$\frac{1}{2} e^{2 y} + C_{1} = 12 (\frac{1}{3} \int \frac{1}{{u_{2}}^{2}} d u_{2})$

Step 2.3.5

Simplify the expression.

Tap for more steps...

Step 2.3.5.1

Simplify.

Tap for more steps...

Step 2.3.5.1.1

Combine $\frac{1}{3}$ and $12$ .

$\frac{1}{2} e^{2 y} + C_{1} = \frac{12}{3} \int \frac{1}{{u_{2}}^{2}} d u_{2}$

Step 2.3.5.1.2

Cancel the common factor of $12$ and $3$ .

Tap for more steps...

Step 2.3.5.1.2.1

Factor $3$ out of $12$ .

$\frac{1}{2} e^{2 y} + C_{1} = \frac{3 \cdot 4}{3} \int \frac{1}{{u_{2}}^{2}} d u_{2}$

Step 2.3.5.1.2.2

Cancel the common factors.

Tap for more steps...

Step 2.3.5.1.2.2.1

Factor $3$ out of $3$ .

$\frac{1}{2} e^{2 y} + C_{1} = \frac{3 \cdot 4}{3 (1)} \int \frac{1}{{u_{2}}^{2}} d u_{2}$

Step 2.3.5.1.2.2.2

Cancel the common factor.

$\frac{1}{2} e^{2 y} + C_{1} = \frac{3 \cdot 4}{3 \cdot 1} \int \frac{1}{{u_{2}}^{2}} d u_{2}$

Step 2.3.5.1.2.2.3

Rewrite the expression.

$\frac{1}{2} e^{2 y} + C_{1} = \frac{4}{1} \int \frac{1}{{u_{2}}^{2}} d u_{2}$

Step 2.3.5.1.2.2.4

Divide $4$ by $1$ .

$\frac{1}{2} e^{2 y} + C_{1} = 4 \int \frac{1}{{u_{2}}^{2}} d u_{2}$

Step 2.3.5.2

Apply basic rules of exponents.

Tap for more steps...

Step 2.3.5.2.1

Move ${u_{2}}^{2}$ out of the denominator by raising it to the $- 1$ power.

$\frac{1}{2} e^{2 y} + C_{1} = 4 \int {({u_{2}}^{2})}^{- 1} d u_{2}$

Step 2.3.5.2.2

Multiply the exponents in ${({u_{2}}^{2})}^{- 1}$ .

Tap for more steps...

Step 2.3.5.2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{1}{2} e^{2 y} + C_{1} = 4 \int {u_{2}}^{2 \cdot - 1} d u_{2}$

Step 2.3.5.2.2.2

Multiply $2$ by $- 1$ .

$\frac{1}{2} e^{2 y} + C_{1} = 4 \int {u_{2}}^{- 2} d u_{2}$

Step 2.3.6

By the Power Rule, the integral of ${u_{2}}^{- 2}$ with respect to $u_{2}$ is $- {u_{2}}^{- 1}$ .

$\frac{1}{2} e^{2 y} + C_{1} = 4 (- {u_{2}}^{- 1} + C_{2})$

Step 2.3.7

Simplify.

Tap for more steps...

Step 2.3.7.1

Rewrite $4 (- {u_{2}}^{- 1} + C_{2})$ as $4 (- \frac{1}{u_{2}}) + C_{2}$ .

$\frac{1}{2} e^{2 y} + C_{1} = 4 (- \frac{1}{u_{2}}) + C_{2}$

Step 2.3.7.2

Simplify.

Tap for more steps...

Step 2.3.7.2.1

Multiply $- 1$ by $4$ .

$\frac{1}{2} e^{2 y} + C_{1} = - 4 \frac{1}{u_{2}} + C_{2}$

Step 2.3.7.2.2

Combine $- 4$ and $\frac{1}{u_{2}}$ .

$\frac{1}{2} e^{2 y} + C_{1} = \frac{- 4}{u_{2}} + C_{2}$

Step 2.3.7.2.3

Move the negative in front of the fraction.

$\frac{1}{2} e^{2 y} + C_{1} = - \frac{4}{u_{2}} + C_{2}$

Step 2.3.8

Replace all occurrences of $u_{2}$ with $2 + 3 x$ .

$\frac{1}{2} e^{2 y} + C_{1} = - \frac{4}{2 + 3 x} + C_{2}$

Step 2.4

Group the constant of integration on the right side as $K$ .

$\frac{1}{2} e^{2 y} = - \frac{4}{2 + 3 x} + K$

Step 3

Solve for $y$ .

Tap for more steps...

Step 3.1

Multiply both sides of the equation by $2$ .

$2 (\frac{1}{2} e^{2 y}) = 2 (- \frac{4}{2 + 3 x} + K)$

Step 3.2

Simplify both sides of the equation.

Tap for more steps...

Step 3.2.1

Simplify the left side.

Tap for more steps...

Step 3.2.1.1

Simplify $2 (\frac{1}{2} e^{2 y})$ .

Tap for more steps...

Step 3.2.1.1.1

Combine $\frac{1}{2}$ and $e^{2 y}$ .

$2 \frac{e^{2 y}}{2} = 2 (- \frac{4}{2 + 3 x} + K)$

Step 3.2.1.1.2

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.2.1.1.2.1

Cancel the common factor.

$2 \frac{e^{2 y}}{2} = 2 (- \frac{4}{2 + 3 x} + K)$

Step 3.2.1.1.2.2

Rewrite the expression.

$e^{2 y} = 2 (- \frac{4}{2 + 3 x} + K)$

Step 3.2.2

Simplify the right side.

Tap for more steps...

Step 3.2.2.1

Simplify $2 (- \frac{4}{2 + 3 x} + K)$ .

Tap for more steps...

Step 3.2.2.1.1

To write $K$ as a fraction with a common denominator, multiply by $\frac{2 + 3 x}{2 + 3 x}$ .

$e^{2 y} = 2 (- \frac{4}{2 + 3 x} + \frac{K (2 + 3 x)}{2 + 3 x})$

Step 3.2.2.1.2

Combine the numerators over the common denominator.

$e^{2 y} = 2 \frac{- 4 + K (2 + 3 x)}{2 + 3 x}$

Step 3.2.2.1.3

Simplify the numerator.

Tap for more steps...

Step 3.2.2.1.3.1

Apply the distributive property.

$e^{2 y} = 2 \frac{- 4 + K \cdot 2 + K (3 x)}{2 + 3 x}$

Step 3.2.2.1.3.2

Move $2$ to the left of $K$ .

$e^{2 y} = 2 \frac{- 4 + 2 \cdot K + K (3 x)}{2 + 3 x}$

Step 3.2.2.1.3.3

Rewrite using the commutative property of multiplication.

$e^{2 y} = 2 \frac{- 4 + 2 K + 3 K x}{2 + 3 x}$

Step 3.2.2.1.4

Simplify terms.

Tap for more steps...

Step 3.2.2.1.4.1

Combine $2$ and $\frac{- 4 + 2 K + 3 K x}{2 + 3 x}$ .

$e^{2 y} = \frac{2 (- 4 + 2 K + 3 K x)}{2 + 3 x}$

Step 3.2.2.1.4.2

Rewrite $- 4$ as $- 1 (4)$ .

$e^{2 y} = \frac{2 (- 1 (4) + 2 K + 3 K x)}{2 + 3 x}$

Step 3.2.2.1.4.3

Factor $- 1$ out of $2 K$ .

$e^{2 y} = \frac{2 (- 1 (4) - (- 2 K) + 3 K x)}{2 + 3 x}$

Step 3.2.2.1.4.4

Factor $- 1$ out of $- 1 (4) - (- 2 K)$ .

$e^{2 y} = \frac{2 (- 1 (4 - 2 K) + 3 K x)}{2 + 3 x}$

Step 3.2.2.1.4.5

Factor $- 1$ out of $3 K x$ .

$e^{2 y} = \frac{2 (- 1 (4 - 2 K) - (- 3 K x))}{2 + 3 x}$

Step 3.2.2.1.4.6

Factor $- 1$ out of $- 1 (4 - 2 K) - (- 3 K x)$ .

$e^{2 y} = \frac{2 (- 1 (4 - 2 K - 3 K x))}{2 + 3 x}$

Step 3.2.2.1.4.7

Move the negative in front of the fraction.

$e^{2 y} = - \frac{2 (4 - 2 K - 3 K x)}{2 + 3 x}$

Step 3.3

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{2 y}) = \ln (- \frac{2 (4 - 2 K - 3 K x)}{2 + 3 x})$

Step 3.4

Expand the left side.

Tap for more steps...

Step 3.4.1

Expand $\ln (e^{2 y})$ by moving $2 y$ outside the logarithm.

$2 y \ln (e) = \ln (- \frac{2 (4 - 2 K - 3 K x)}{2 + 3 x})$

Step 3.4.2

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

$2 y \cdot 1 = \ln (- \frac{2 (4 - 2 K - 3 K x)}{2 + 3 x})$

Step 3.4.3

Multiply $2$ by $1$ .

$2 y = \ln (- \frac{2 (4 - 2 K - 3 K x)}{2 + 3 x})$

Step 3.5

Divide each term in $2 y = \ln (- \frac{2 (4 - 2 K - 3 K x)}{2 + 3 x})$ by $2$ and simplify.

Tap for more steps...

Step 3.5.1

Divide each term in $2 y = \ln (- \frac{2 (4 - 2 K - 3 K x)}{2 + 3 x})$ by $2$ .

$\frac{2 y}{2} = \frac{\ln (- \frac{2 (4 - 2 K - 3 K x)}{2 + 3 x})}{2}$

Step 3.5.2

Simplify the left side.

Tap for more steps...

Step 3.5.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.5.2.1.1

Cancel the common factor.

$\frac{2 y}{2} = \frac{\ln (- \frac{2 (4 - 2 K - 3 K x)}{2 + 3 x})}{2}$

Step 3.5.2.1.2

Divide $y$ by $1$ .

$y = \frac{\ln (- \frac{2 (4 - 2 K - 3 K x)}{2 + 3 x})}{2}$

Step 4

Simplify the constant of integration.

$y = \frac{\ln (- \frac{2 (K + K x)}{2 + 3 x})}{2}$

Step 5

Use the initial condition to find the value of $K$ by substituting $- 2$ for $x$ and $0$ for $y$ in $y = \frac{\ln (- \frac{2 (K + K x)}{2 + 3 x})}{2}$ .

$0 = \frac{\ln (- \frac{2 (K + K \cdot - 2)}{2 + 3 \cdot - 2})}{2}$

Step 6

Solve for $K$ .

Tap for more steps...

Step 6.1

Set the numerator equal to zero.

$\ln (- \frac{2 (K + K \cdot - 2)}{2 + 3 \cdot - 2}) = 0$

Step 6.2

Solve the equation for $K$ .

Tap for more steps...

Step 6.2.1

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

$e^{\ln (- \frac{2 (K + K \cdot - 2)}{2 + 3 \cdot - 2})} = e^{0}$

Step 6.2.2

Rewrite $\ln (- \frac{2 (K + K \cdot - 2)}{2 + 3 \cdot - 2}) = 0$ 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 $b^{y} = x$ .

$e^{0} = - \frac{2 (K + K \cdot - 2)}{2 + 3 \cdot - 2}$

Step 6.2.3

Solve for $K$ .

Tap for more steps...

Step 6.2.3.1

Rewrite the equation as $- \frac{2 (K + K \cdot - 2)}{2 + 3 \cdot - 2} = e^{0}$ .

$- \frac{2 (K + K \cdot - 2)}{2 + 3 \cdot - 2} = e^{0}$

Step 6.2.3.2

Divide each term in $- \frac{2 (K + K \cdot - 2)}{2 + 3 \cdot - 2} = e^{0}$ by $- 1$ and simplify.

Tap for more steps...

Step 6.2.3.2.1

Divide each term in $- \frac{2 (K + K \cdot - 2)}{2 + 3 \cdot - 2} = e^{0}$ by $- 1$ .

$\frac{- \frac{2 (K + K \cdot - 2)}{2 + 3 \cdot - 2}}{- 1} = \frac{e^{0}}{- 1}$

Step 6.2.3.2.2

Simplify the left side.

Tap for more steps...

Step 6.2.3.2.2.1

Reduce the expression by cancelling the common factors.

Tap for more steps...

Step 6.2.3.2.2.1.1

Dividing two negative values results in a positive value.

$\frac{\frac{2 (K + K \cdot - 2)}{2 + 3 \cdot - 2}}{1} = \frac{e^{0}}{- 1}$

Step 6.2.3.2.2.1.2

Divide $\frac{2 (K + K \cdot - 2)}{2 + 3 \cdot - 2}$ by $1$ .

$\frac{2 (K + K \cdot - 2)}{2 + 3 \cdot - 2} = \frac{e^{0}}{- 1}$

Step 6.2.3.2.2.1.3

Cancel the common factor of $2$ and $2 + 3 \cdot - 2$ .

Tap for more steps...

Step 6.2.3.2.2.1.3.1

Reorder terms.

$\frac{2 (K + K \cdot - 2)}{2 - 2 \cdot 3} = \frac{e^{0}}{- 1}$

Step 6.2.3.2.2.1.3.2

Cancel the common factors.

Tap for more steps...

Step 6.2.3.2.2.1.3.2.1

Factor $2$ out of $2$ .

$\frac{2 (K + K \cdot - 2)}{2 (1) - 2 \cdot 3} = \frac{e^{0}}{- 1}$

Step 6.2.3.2.2.1.3.2.2

Factor $2$ out of $- 2 \cdot 3$ .

$\frac{2 (K + K \cdot - 2)}{2 (1) + 2 (- 1 \cdot 3)} = \frac{e^{0}}{- 1}$

Step 6.2.3.2.2.1.3.2.3

Factor $2$ out of $2 (1) + 2 (- 1 \cdot 3)$ .

$\frac{2 (K + K \cdot - 2)}{2 (1 - 1 \cdot 3)} = \frac{e^{0}}{- 1}$

Step 6.2.3.2.2.1.3.2.4

Cancel the common factor.

$\frac{2 (K + K \cdot - 2)}{2 (1 - 1 \cdot 3)} = \frac{e^{0}}{- 1}$

Step 6.2.3.2.2.1.3.2.5

Rewrite the expression.

$\frac{K + K \cdot - 2}{1 - 1 \cdot 3} = \frac{e^{0}}{- 1}$

Step 6.2.3.2.2.2

Simplify the numerator.

Tap for more steps...

Step 6.2.3.2.2.2.1

Move $- 2$ to the left of $K$ .

$\frac{K - 2 \cdot K}{1 - 1 \cdot 3} = \frac{e^{0}}{- 1}$

Step 6.2.3.2.2.2.2

Subtract $2 K$ from $K$ .

$\frac{- K}{1 - 1 \cdot 3} = \frac{e^{0}}{- 1}$

Step 6.2.3.2.2.3

Simplify the denominator.

Tap for more steps...

Step 6.2.3.2.2.3.1

Multiply $- 1$ by $3$ .

$\frac{- K}{1 - 3} = \frac{e^{0}}{- 1}$

Step 6.2.3.2.2.3.2

Subtract $3$ from $1$ .

$\frac{- K}{- 2} = \frac{e^{0}}{- 1}$

Step 6.2.3.2.2.4

Dividing two negative values results in a positive value.

$\frac{K}{2} = \frac{e^{0}}{- 1}$

Step 6.2.3.2.3

Simplify the right side.

Tap for more steps...

Step 6.2.3.2.3.1

Move the negative one from the denominator of $\frac{e^{0}}{- 1}$ .

$\frac{K}{2} = - 1 \cdot e^{0}$

Step 6.2.3.2.3.2

Rewrite $- 1 \cdot e^{0}$ as $- e^{0}$ .

$\frac{K}{2} = - e^{0}$

Step 6.2.3.2.3.3

Anything raised to $0$ is $1$ .

$\frac{K}{2} = - 1 \cdot 1$

Step 6.2.3.2.3.4

Multiply $- 1$ by $1$ .

$\frac{K}{2} = - 1$

Step 6.2.3.3

Multiply both sides of the equation by $2$ .

$2 \frac{K}{2} = 2 \cdot - 1$

Step 6.2.3.4

Simplify both sides of the equation.

Tap for more steps...

Step 6.2.3.4.1

Simplify the left side.

Tap for more steps...

Step 6.2.3.4.1.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 6.2.3.4.1.1.1

Cancel the common factor.

$2 \frac{K}{2} = 2 \cdot - 1$

Step 6.2.3.4.1.1.2

Rewrite the expression.

$K = 2 \cdot - 1$

Step 6.2.3.4.2

Simplify the right side.

Tap for more steps...

Step 6.2.3.4.2.1

Multiply $2$ by $- 1$ .

$K = - 2$

Step 7

Substitute $- 2$ for $K$ in $y = \frac{\ln (- \frac{2 (K + K x)}{2 + 3 x})}{2}$ and simplify.

Tap for more steps...

Step 7.1

Substitute $- 2$ for $K$ .

$y = \frac{\ln (- \frac{2 (- 2 + K x)}{2 + 3 x})}{2}$

Step 7.2

Rewrite $\frac{\ln (- \frac{2 (- 2 + K x)}{2 + 3 x})}{2}$ as $\frac{1}{2} \ln (- \frac{2 (- 2 + K x)}{2 + 3 x})$ .

$y = \frac{1}{2} \ln (- \frac{2 (- 2 + K x)}{2 + 3 x})$

Step 7.3

Simplify $\frac{1}{2} \ln (- \frac{2 (- 2 + K x)}{2 + 3 x})$ by moving $\frac{1}{2}$ inside the logarithm.

$y = \ln ({(- \frac{2 (- 2 + K x)}{2 + 3 x})}^{\frac{1}{2}})$

Step 7.4

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Tap for more steps...

Step 7.4.1

Apply the product rule to $- \frac{2 (- 2 + K x)}{2 + 3 x}$ .

$y = \ln ({(- 1)}^{\frac{1}{2}} {(\frac{2 (- 2 + K x)}{2 + 3 x})}^{\frac{1}{2}})$

Step 7.4.2

Apply the product rule to $\frac{2 (- 2 + K x)}{2 + 3 x}$ .

$y = \ln ({(- 1)}^{\frac{1}{2}} \frac{{(2 (- 2 + K x))}^{\frac{1}{2}}}{{(2 + 3 x)}^{\frac{1}{2}}})$

Step 7.4.3

Apply the product rule to $2 (- 2 + K x)$ .

$y = \ln ({(- 1)}^{\frac{1}{2}} \frac{2^{\frac{1}{2}} {(- 2 + K x)}^{\frac{1}{2}}}{{(2 + 3 x)}^{\frac{1}{2}}})$

Step 7.5

Rewrite ${(- 1)}^{\frac{1}{2}}$ as $\sqrt{{(- 1)}^{1}}$ .

$y = \ln (\sqrt{{(- 1)}^{1}} \frac{2^{\frac{1}{2}} {(- 2 + K x)}^{\frac{1}{2}}}{{(2 + 3 x)}^{\frac{1}{2}}})$

Step 7.6

Evaluate the exponent.

$y = \ln (\sqrt{- 1} \frac{2^{\frac{1}{2}} {(- 2 + K x)}^{\frac{1}{2}}}{{(2 + 3 x)}^{\frac{1}{2}}})$

Step 7.7

Rewrite $\sqrt{- 1}$ as $i$ .

$y = \ln (i \frac{2^{\frac{1}{2}} {(- 2 + K x)}^{\frac{1}{2}}}{{(2 + 3 x)}^{\frac{1}{2}}})$

Step 7.8

Combine $i$ and $\frac{2^{\frac{1}{2}} {(- 2 + K x)}^{\frac{1}{2}}}{{(2 + 3 x)}^{\frac{1}{2}}}$ .

$y = \ln (\frac{i \cdot 2^{\frac{1}{2}} {(- 2 + K x)}^{\frac{1}{2}}}{{(2 + 3 x)}^{\frac{1}{2}}})$