Calculus Examples

$e^{4 x + 3} = 9$

Step 1

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

$\ln (e^{4 x + 3}) = \ln (9)$

Step 2

Expand the left side.

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Step 2.1

Expand $\ln (e^{4 x + 3})$ by moving $4 x + 3$ outside the logarithm.

$(4 x + 3) \ln (e) = \ln (9)$

Step 2.2

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

$(4 x + 3) \cdot 1 = \ln (9)$

Step 2.3

Multiply $4 x + 3$ by $1$ .

$4 x + 3 = \ln (9)$

Step 3

Subtract $3$ from both sides of the equation.

$4 x = \ln (9) - 3$

Step 4

Divide each term in $4 x = \ln (9) - 3$ by $4$ and simplify.

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Step 4.1

Divide each term in $4 x = \ln (9) - 3$ by $4$ .

$\frac{4 x}{4} = \frac{\ln (9)}{4} + \frac{- 3}{4}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $4$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{\ln (9)}{4} + \frac{- 3}{4}$

Step 4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (9)}{4} + \frac{- 3}{4}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Simplify each term.

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Step 4.3.1.1

Rewrite $\ln (9)$ as $\ln (3^{2})$ .

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

Step 4.3.1.2

Expand $\ln (3^{2})$ by moving $2$ outside the logarithm.

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

Step 4.3.1.3

Cancel the common factor of $2$ and $4$ .

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Step 4.3.1.3.1

Factor $2$ out of $2 \ln (3)$ .

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

Step 4.3.1.3.2

Cancel the common factors.

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Step 4.3.1.3.2.1

Factor $2$ out of $4$ .

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

Step 4.3.1.3.2.2

Cancel the common factor.

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

Step 4.3.1.3.2.3

Rewrite the expression.

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

Step 4.3.1.4

Move the negative in front of the fraction.

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

Step 5

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = - 0.20069385 \dots$