Precalculus Examples

${(e^{2 \ln (2) - 3 \ln (2) + 1})}^{2}$

Step 1

Simplify each term.

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Step 1.1

Simplify $2 \ln (2)$ by moving $2$ inside the logarithm.

${(e^{\ln (2^{2}) - 3 \ln (2) + 1})}^{2}$

Step 1.2

Raise $2$ to the power of $2$ .

${(e^{\ln (4) - 3 \ln (2) + 1})}^{2}$

Step 1.3

Simplify $- 3 \ln (2)$ by moving $3$ inside the logarithm.

${(e^{\ln (4) - \ln (2^{3}) + 1})}^{2}$

Step 1.4

Raise $2$ to the power of $3$ .

${(e^{\ln (4) - \ln (8) + 1})}^{2}$

Step 2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

${(e^{\ln (\frac{4}{8}) + 1})}^{2}$

Step 3

Cancel the common factor of $4$ and $8$ .

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Step 3.1

Factor $4$ out of $4$ .

${(e^{\ln (\frac{4 (1)}{8}) + 1})}^{2}$

Step 3.2

Cancel the common factors.

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Step 3.2.1

Factor $4$ out of $8$ .

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

Step 3.2.2

Cancel the common factor.

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

Step 3.2.3

Rewrite the expression.

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

Step 4

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

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Step 4.1

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

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

Step 4.2

Apply the distributive property.

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

Step 4.3

Multiply $\ln (\frac{1}{2}) \cdot 2$ .

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Step 4.3.1

Reorder $\ln (\frac{1}{2})$ and $2$ .

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

Step 4.3.2

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

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

Step 4.4

Multiply $2$ by $1$ .

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

Step 4.5

Simplify each term.

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Step 4.5.1

Apply the product rule to $\frac{1}{2}$ .

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

Step 4.5.2

One to any power is one.

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

Step 4.5.3

Raise $2$ to the power of $2$ .

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

Step 5

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$1.84726402 \dots$