# #1 Makamae Aquino Period 2A EXPONENT LAWS. 6 LAWS OF EXPONENTS - Negative Expo - First Power - Zero Power - MA - DS - PM.

## Presentation on theme: "#1 Makamae Aquino Period 2A EXPONENT LAWS. 6 LAWS OF EXPONENTS - Negative Expo - First Power - Zero Power - MA - DS - PM."— Presentation transcript:

#1 Makamae Aquino Period 2A EXPONENT LAWS

6 LAWS OF EXPONENTS - Negative Expo - First Power - Zero Power - MA - DS - PM

NEGATIVE EXPONENT LAW -Negative exponents express small numbers. -Law: Take base of negative exponent, flip its location (numerator/denominator) & take away negative sign of exponent.

FIRST POWER LAW -When a number is raised to the power of one. -Law: when a number is raised to the power of one, the number is itself.

ZERO POWER LAW -When a number is raised to the power of zero. -Law: when a number is raised to the power of zero, it is one.

MADSPM MA= Multiply same base & Add expo DS= Divide same base & Subtract expo PM= Power of a power & Multiply expo

MA LAW -Used when multiplying -Law: When multiplying exponent numbers with the same base, add exponent.

STEP ONE- COEFFICIENTS & BASES Multiply the coefficients (if coefficients are the same, just drop them). Drop the base/bases (variable). Problem: 5 x to the fourth multiplied by 3 x to the third. 5 multiplied by 3 is 15. Bring down the x.

STEP TWO- EXPONENTS Problem: 5 x to the fourth multiplied by 3 x to the third. Add the exponents (ex. x to the sixth and x to the eighth = x6+8). x4+3=x7

STEP THREE- SIMPLIFY Problem: 5 x to the fourth multiplied by 3 x to the third. Put everything together. Final Answer: 15x7

DS LAW -Used when dividing -Law: Dividing same base number, subtract exponent (subtract bottom from top)

STEP ONE- COEFFICIENTS & BASES Problem: 4 to the 5 th power divided by 12 squared. -Drop the coefficients and bases one by one by one (if the coefficients are the same, drop the coefficients. If the coefficients are not the same, simplify them [keep # on top, on the top and # on bottom, on the bottom]) Drop the base (variables). Keep the base on top, on the top, and base on bottom, on the bottom. 4 and 12 is simplified to 1 and 3

STEP TWO- EXPONENTS Problem: 4 to the 5 th power divided by 12 squared Bring the exponents down to the numerator (1) and subtract them (5 - 2). New Answer: 1 to 5-2 divided by 3

STEP THREE- SIMPLIFY Problem: 4 to the 5 th power divided by 12 squared Simplify the new answer (1 to the 5-2 divided by 3). Final answer: 1 (numerator) divided by 3 (denominator)= 3 5-2=3. 1x1x1=1

PM LAW -Used when figuring out a power of a power. -Law: Taking a power of a power, and multiplying the exponents.

STEP ONE- MULTIPLYING COEFFICIENTS Problem: (-2 multiplied by x to the seventh) to the fourth. Drop the coefficient down and multiply. *Everything in the () needs to be multiplied. If the number has no exponent, multiply the coefficient by the exponent. If there is an exponent with the coefficient, multiply the exponent by the exponent outside the (). -2 multiplied by four=16. (The exponent is even, therefore the coefficient becomes positive).

STEP TWO- MULTIPLYING EXPONENTS Problem: (-2 multiplied by x to the seventh) to the fourth. Drop the variables and their exponents. Multiply all the variable’s exponents by the exponent outside the (). x to the seventh multiplied by 4=28

STEP THREE- SIMPLIFY Problem: (-2 multiplied by x to the seventh) to the fourth. Put everything together for the final answer. Final answer: 16 and x28= 16x to the 28.

Similar presentations