Calculus Examples

$1 = \frac{1}{4} \ln (4 e^{2}) + c$

Step 1

Rewrite the equation as $\frac{1}{4} \cdot \ln (4 e^{2}) + c = 1$ .

$\frac{1}{4} \cdot \ln (4 e^{2}) + c = 1$

Step 2

Simplify each term.

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Step 2.1

Simplify $\frac{1}{4} \ln (4 e^{2})$ by moving $\frac{1}{4}$ inside the logarithm.

$\ln ({(4 e^{2})}^{\frac{1}{4}}) + c = 1$

Step 2.2

Apply the product rule to $4 e^{2}$ .

$\ln (4^{\frac{1}{4}} {(e^{2})}^{\frac{1}{4}}) + c = 1$

Step 2.3

Multiply the exponents in ${(e^{2})}^{\frac{1}{4}}$ .

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Step 2.3.1

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

$\ln (4^{\frac{1}{4}} e^{2 (\frac{1}{4})}) + c = 1$

Step 2.3.2

Cancel the common factor of $2$ .

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Step 2.3.2.1

Factor $2$ out of $4$ .

$\ln (4^{\frac{1}{4}} e^{2 \frac{1}{2 (2)}}) + c = 1$

Step 2.3.2.2

Cancel the common factor.

$\ln (4^{\frac{1}{4}} e^{2 \frac{1}{2 \cdot 2}}) + c = 1$

Step 2.3.2.3

Rewrite the expression.

$\ln (4^{\frac{1}{4}} e^{\frac{1}{2}}) + c = 1$

Step 3

Subtract $\ln (4^{\frac{1}{4}} e^{\frac{1}{2}})$ from both sides of the equation.

$c = 1 - \ln (4^{\frac{1}{4}} e^{\frac{1}{2}})$

Step 4

The result can be shown in multiple forms.

Exact Form:

$c = 1 - \ln (4^{\frac{1}{4}} e^{\frac{1}{2}})$

Decimal Form:

$c = 0.15342640 \dots$