Precalculus Examples

$3000 = p {(1 + \frac{0.525}{2})}^{12}$

Step 1

Rewrite the equation as $p {(1 + \frac{0.525}{2})}^{12} = 3000$ .

$p {(1 + \frac{0.525}{2})}^{12} = 3000$

Step 2

Simplify.

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Step 2.1

Divide $0.525$ by $2$ .

$p {(1 + 0.2625)}^{12} = 3000$

Step 2.2

Add $1$ and $0.2625$ .

$p \cdot {1.2625}^{12} = 3000$

Step 3

Divide each term in $p \cdot {1.2625}^{12} = 3000$ by ${1.2625}^{12}$ and simplify.

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Step 3.1

Divide each term in $p \cdot {1.2625}^{12} = 3000$ by ${1.2625}^{12}$ .

$\frac{p \cdot {1.2625}^{12}}{{1.2625}^{12}} = \frac{3000}{{1.2625}^{12}}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of ${1.2625}^{12}$ .

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Step 3.2.1.1

Cancel the common factor.

$\frac{p \cdot {1.2625}^{12}}{{1.2625}^{12}} = \frac{3000}{{1.2625}^{12}}$

Step 3.2.1.2

Divide $p$ by $1$ .

$p = \frac{3000}{{1.2625}^{12}}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Raise $1.2625$ to the power of $12$ .

$p = \frac{3000}{16.39746231}$

Step 3.3.2

Divide $3000$ by $16.39746231$ .

$p = 182.95513916$