Exponents:
When a number or variable is multiplied with itself a number of times, it gives rise to the exponential value of that number or variable. That is to say, if we have a * a * a * a, we write that as a^4, where 4 is the exponent and a is the base. We use 4 here because a is multiplied to itself 4 times. In general if a is multiplied to itself n times like this: a*a*a*a*a….. n times, then that would be same as writing a^n. Where n = exponent and a = base. So, 2^3 = 2*2*2 = *8; or 3^2 = 3*3 = 9 etc.
Example:
A sample of bacteria doubles every hour. If the initial number is 100, how many bacteria would be there after 8 hours?
Solution: The above problem pertains to exponential growth. So after one hour, the number of bacteria would be = 100 * 2. After another one hour it would be = 100 * 2 * 2 = 100 * 2^2. After the third hour the number would be = 100 * 2^2 * 2 = 100 * 2^3 and so on. So after 8 hours the number would be = 100 * 2^8 = 100 * 256 = 25600.
Negative exponents:
Exponents can also be negative. A negative exponent would mean the positive exponent of the reciprocal of the base. Therefore a^(-n) = (1/a)^n (or 1/a^n ). Therefore, 5^(-2) = (1/5)^2 = 1^2/5^2 = 1/25, or 4^(-3) = (1/4)^3 = 1^3/4^3 = 1/64.
Multiplying exponents:
Instead of multiplying a*a*a*a…. n times, if we are asked to multiply (a^m) * (a^m) * (a^m) * (a^m) …… n times, then the answer would be (a^m)^n = (a*a*a*a…. m times)^n = (a*a*a*a…. m times) * (a*a*a*a…. m times) * (a*a*a*a…. m times) * (a*a*a*a…. m times) ….. n times = a*a*a*a….. (m*n) times. Therefore (a^m)^n = a^(mn). For example, (11^2)^3 = 11^(2*3) = 11^6, because 11^2 = 11*11, and (11^2)^3 = (11*11)^3 = (11*11) * (11*11) * (11*11) = 11^6
Extending the above concept further, if we have ((a^m)^n)^p = a^(mnp). For example, ((5^2)^2)^2 = 5^(2*2*2) = 5^8.
Example: A corporate deposit is such that $500 invested doubles every 5 years. What would be the amount due after 25 years?
Solution: Amount after the 1st 5 years = 500*2 = 1000
Amount after the 2nd 5 years = 500*2 * 2 = 2000
Therefore the amount after 5th years = 500 * 2^5 = 500 * 32 = 16000.
When a number or variable is multiplied with itself a number of times, it gives rise to the exponential value of that number or variable. That is to say, if we have a * a * a * a, we write that as a^4, where 4 is the exponent and a is the base. We use 4 here because a is multiplied to itself 4 times. In general if a is multiplied to itself n times like this: a*a*a*a*a….. n times, then that would be same as writing a^n. Where n = exponent and a = base. So, 2^3 = 2*2*2 = *8; or 3^2 = 3*3 = 9 etc.
Example:
A sample of bacteria doubles every hour. If the initial number is 100, how many bacteria would be there after 8 hours?
Solution: The above problem pertains to exponential growth. So after one hour, the number of bacteria would be = 100 * 2. After another one hour it would be = 100 * 2 * 2 = 100 * 2^2. After the third hour the number would be = 100 * 2^2 * 2 = 100 * 2^3 and so on. So after 8 hours the number would be = 100 * 2^8 = 100 * 256 = 25600.
Negative exponents:
Exponents can also be negative. A negative exponent would mean the positive exponent of the reciprocal of the base. Therefore a^(-n) = (1/a)^n (or 1/a^n ). Therefore, 5^(-2) = (1/5)^2 = 1^2/5^2 = 1/25, or 4^(-3) = (1/4)^3 = 1^3/4^3 = 1/64.
Multiplying exponents:
Instead of multiplying a*a*a*a…. n times, if we are asked to multiply (a^m) * (a^m) * (a^m) * (a^m) …… n times, then the answer would be (a^m)^n = (a*a*a*a…. m times)^n = (a*a*a*a…. m times) * (a*a*a*a…. m times) * (a*a*a*a…. m times) * (a*a*a*a…. m times) ….. n times = a*a*a*a….. (m*n) times. Therefore (a^m)^n = a^(mn). For example, (11^2)^3 = 11^(2*3) = 11^6, because 11^2 = 11*11, and (11^2)^3 = (11*11)^3 = (11*11) * (11*11) * (11*11) = 11^6
Extending the above concept further, if we have ((a^m)^n)^p = a^(mnp). For example, ((5^2)^2)^2 = 5^(2*2*2) = 5^8.
Example: A corporate deposit is such that $500 invested doubles every 5 years. What would be the amount due after 25 years?
Solution: Amount after the 1st 5 years = 500*2 = 1000
Amount after the 2nd 5 years = 500*2 * 2 = 2000
Therefore the amount after 5th years = 500 * 2^5 = 500 * 32 = 16000.
