Finite Math Examples

$90000 = A \cdot {(1 + (\frac{0.03}{12}))}^{- 12 \cdot 51}$

Step 1

Rewrite the equation as $A \cdot {(1 + \frac{0.03}{12})}^{- 12 \cdot 51} = 90000$ .

$A \cdot {(1 + \frac{0.03}{12})}^{- 12 \cdot 51} = 90000$

Step 2

Simplify.

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Step 2.1

Multiply $- 12$ by $51$ .

$A \cdot {(1 + \frac{0.03}{12})}^{- 612} = 90000$

Step 2.2

Divide $0.03$ by $12$ .

$A \cdot {(1 + 0.0025)}^{- 612} = 90000$

Step 2.3

Add $1$ and $0.0025$ .

$A \cdot {1.0025}^{- 612} = 90000$

Step 3

Divide each term in $A \cdot {1.0025}^{- 612} = 90000$ by ${1.0025}^{- 612}$ and simplify.

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Step 3.1

Divide each term in $A \cdot {1.0025}^{- 612} = 90000$ by ${1.0025}^{- 612}$ .

$\frac{A \cdot {1.0025}^{- 612}}{{1.0025}^{- 612}} = \frac{90000}{{1.0025}^{- 612}}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of ${1.0025}^{- 612}$ .

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Step 3.2.1.1

Cancel the common factor.

$\frac{A \cdot {1.0025}^{- 612}}{{1.0025}^{- 612}} = \frac{90000}{{1.0025}^{- 612}}$

Step 3.2.1.2

Divide $A$ by $1$ .

$A = \frac{90000}{{1.0025}^{- 612}}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Move ${1.0025}^{- 612}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$A = 90000 \cdot {1.0025}^{612}$

Step 3.3.2

Raise $1.0025$ to the power of $612$ .

$A = 90000 \cdot 4.60936766$

Step 3.3.3

Multiply $90000$ by $4.60936766$ .

$A = 414843.08979959$