Raises a number to a power. For example, you can use this function to square the number 2 to get the result of 4.


NewNumber := POWER(Number, Power)

Parameters

Number

Type: Integer

The number you want to raise exponentially. This number is the base in the exponential function.

Power

Type: Integer

The exponent in the exponential function.

Property Value/Return Value

Example

This example requires that you create the following variables and text constants in the C/AL Globals window.

Variable name	DataType
Number1	Decimal
Power1	Decimal
Result1	Decimal

Text constant name	Constant value
Text000	%1 raised to the power of %2 = %3

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Number1 := 64; Power1 := 0.5; Result1 := POWER(Number1, Power1); MESSAGE(Text000, Number1, Power1, Result1)

On a computer with the regional format set to English (United States), the first message window displays the following:

64 raised to the power of 0.5 = 8

This example shows that raising a number to the power of 0.5 corresponds to the square root of the number.

This example shows a typical use for the POWER function.

If a principal amount P is deposited at interest rate R and compounded annually, then at the end of N years, the accumulated amount (A) is:

A = P(1 + R)^N

For example, you place LCY 2800 into a bank account that pays 5 percent, which is compounded quarterly. To determine what the amount will be in eight years, you must consider:

N = 32 payment periods (8 years times 4 quarterly periods)

R = 0.0125 per period (5 percent divided by 4 quarterly periods)

The accumulated amount A is:

A = LCY 2800(1 + 0.0125)³² =LCY 2800(1.4881) = LCY 4166.77

If a principal amount P is deposited at the end of each year at interest rate R (in decimal notation) compounded annually, then at the end of N years, the accumulated amount is:

A = P[((1 + R)^N - 1)/R]

This is typically called an annuity.

For example, you have an annuity in which a payment of LCY 500 is made at the end of each year. The interest on this annuity is 4 percent, which is compounded annually. To determine what the annuity will be worth in 20 years, you must consider:

R = 0.04

N = 20

The amount of the annuity A will be:

A = LCY 500[((1 + 0.04)²⁰ - 1)/0.04 = LCY 14,889.04

This example requires that you create the following variables and text constants in the C/AL Globals window.

Variable name	DataType
P	Decimal
R	Decimal
N	Decimal
A	Decimal
FormatString	Text

Text constant name	Constant value
Text000	Principal $%1 at a 5 percent interest rate is compounded quarterly.\
Text001	(Rate = %2)\
Text002	The amount after %3 years = $%4.
Text003	Principal $%1 is deposited at the end of each year at a 4 percent interest rate, compounded annually.\
Text004	(Rate = %2)\
Text005	The amount after %3 years = $%4.

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FormatString := '<Precision,2><Standard Format,1>'; // Example 1 P := 2800; R := 0.0125; N := 32; A = P * (POWER(1 + R, N)); MESSAGE(Text000 + Text001 + Text002, P, R, N, FORMAT(A,0,FormatString); // Example 2 P = 500; R = 0.04; N = 20; A = P * ((POWER(1 + R, N) - 1)/R); MESSAGE(Text001, P, R, N, FORMAT(A,0,FormatString));

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FormatString := '<Precision,2><Standard Format,1>';
// Example 1
P := 2800;
R := 0.0125;
N := 32;
A = P * (POWER(1 + R, N));
MESSAGE(Text000 + Text001 + Text002, P, R, N, FORMAT(A,0,FormatString);
// Example 2
P = 500;
R = 0.04;
N = 20;
A = P * ((POWER(1 + R, N) - 1)/R);
MESSAGE(Text001, P, R, N, FORMAT(A,0,FormatString));

On a computer with the regional format set to English (United States), the first message window displays the following:

Principal $2,800 at a 5 percent interest rate is compounded quarterly.

(Rate = 0.0125)

The amount after 32 years = $4166.77.

The second message window displays the following:

Principal $500 is deposited at the end of each year at a 4 percent interest rate, compounded annually.

(Rate = 0.04)

The amount after 20 years = $14889.04.

Parameters

Property Value/Return Value

Example

See Also

Reference