This post is a part of [
GRE MATH BOOK]
Frequency of the concepts tested:
HighEXPONENTS
Exponents are a "shortcut" method of showing a number that was multiplied by itself several times. For instance, number \(a\) multiplied \(n\) times can be written as \(a^n\), where \(a\) represents the base, the number that is multiplied by itself \(n\) times and \(n\) represents the exponent. The exponent indicates how many times to multiple the base, \(a\), by itself.
Exponents one and zero:
\(a^0=1\) Any nonzero number to the power of 0 is 1.
For example: \(5^0=1\) and \((-3)^0=1\)
• Note: the case of 0^0 is not tested on the GRE.
\(a^1=a\) Any number to the power 1 is itself.
Powers of zero:
If the exponent is positive, the power of zero is zero: \(0^n = 0\), where \(n > 0\).
If the exponent is negative, the power of zero (\(0^n\), where \(n < 0\)) is undefined, because division by zero is implied.
Powers of one:
\(1^n=1\) The integer powers of one are one.
Negative powers:
\(a^{-n}=\frac{1}{a^n}\)
Powers of minus one:
If n is an even integer, then \((-1)^n=1\).
If n is an odd integer, then \((-1)^n =-1\).
Operations involving the same exponents:
Keep the exponent, multiply or divide the bases
\(a^n*b^n=(ab)^n\)
\(\frac{a^n}{b^n}=(\frac{a}{b})^n\)
\((a^m)^n=a^{mn}\)
\(a^{m^n}=a^{(m^n)}\) and not \((a^m)^n\)
Operations involving the same bases:
Keep the base, add or subtract the exponent (add for multiplication, subtract for division)
\(a^n*a^m=a^{n+m}\)
\(\frac{a^n}{a^m}=a^{n-m}\)
Fraction as power:
\(a^{\frac{1}{n}}=\sqrt[n]{a}\)
\(a^{\frac{m}{n}}=\sqrt[n]{a^m}\)
Exponential Equations:
When solving equations with even exponents, we must consider both positive and negative possibilities for the solutions.
For instance \(a^2=25\), the two possible solutions are \(5\) and \(-5\).
When solving equations with odd exponents, we'll have only one solution.
For instance for \(a^3=8\), solution is \(a=2\) and for \(a^3=-8\), solution is \(a=-2\).
Exponents and divisibility:
\(a^n-b^n\) is ALWAYS divisible by \(a-b\).
\(a^n-b^n\) is divisible by \(a+b\) if \(n\) is even.
\(a^n + b^n\) is divisible by \(a+b\) if \(n\) is odd, and not divisible by a+b if n is even.
LAST DIGIT OF A PRODUCT
Last \(n\) digits of a product of integers are last \(n\) digits of the product of last \(n\) digits of these integers.
For instance last 2 digits of 845*9512*408*613 would be the last 2 digits of 45*12*8*13=540*104=40*4=160=60
Example: The last digit of 85945*89*58307=5*9*7=45*7=35=5?
LAST DIGIT OF A POWER
Determining the last digit of \((xyz)^n\):
1. Last digit of \((xyz)^n\) is the same as that of \(z^n\);
2. Determine the cyclicity number \(c\) of \(z\);
3. Find the remainder \(r\) when \(n\) divided by the cyclisity;
4. When \(r>0\), then last digit of \((xyz)^n\) is the same as that of \(z^r\) and when \(r=0\), then last digit of \((xyz)^n\) is the same as that of \(z^c\), where \(c\) is the cyclisity number.
• Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.
• Integers ending with 2, 3, 7 and 8 have a cyclicity of 4.
• Integers ending with 4 (eg. \((xy4)^n\)) have a cyclisity of 2. When n is odd \((xy4)^n\) will end with 4 and when n is even \((xy4)^n\) will end with 6.
• Integers ending with 9 (eg. \((xy9)^n\)) have a cyclisity of 2. When n is odd \((xy9)^n\) will end with 9 and when n is even \((xy9)^n\) will end with 1.
Example: What is the last digit of \(127^{39}\)?
Solution: Last digit of \(127^{39}\) is the same as that of \(7^{39}\). Now we should determine the cyclisity of \(7\):
1. 7^1=7 (last digit is 7)
2. 7^2=9 (last digit is 9)
3. 7^3=3 (last digit is 3)
4. 7^4=1 (last digit is 1)
5. 7^5=7 (last digit is 7 again!)
...
So, the cyclisity of 7 is 4.
Now divide 39 (power) by 4 (cyclisity), remainder is 3.So, the last digit of \(127^{39}\) is the same as that of the last digit of \(7^{39}\), is the same as that of the last digit of \(7^3\), which is \(3\).
ROOTS
Roots (or radicals) are the "opposite" operation of applying exponents. For instance x^2=16 and square root of 16=4.
General rules:
• \(\sqrt{x}\sqrt{y}=\sqrt{xy}\) and \(\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}\).
• \((\sqrt{x})^n=\sqrt{x^n}\)
• \(x^{\frac{1}{n}}=\sqrt[n]{x}\)
• \(x^{\frac{n}{m}}=\sqrt[m]{x^n}\)
• \({\sqrt{a}}+{\sqrt{b}}\neq{\sqrt{a+b}}\)
• \(\sqrt{x^2}=|x|\), when \(x\leq{0}\), then \(\sqrt{x^2}=-x\) and when \(x\geq{0}\), then \(\sqrt{x^2}=x\)
• When the GRE provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.
That is, \(\sqrt{25}=5\), NOT +5 or -5. In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5. Even roots have only a positive value on the GRE.
• Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).
• For GRE it's good to memorize following values:
\(\sqrt{2}\approx{1.41}\)
\(\sqrt{3}\approx{1.73}\)
\(\sqrt{5}\approx{2.24}\)
\(\sqrt{6}\approx{2.45}\)
\(\sqrt{7}\approx{2.65}\)
\(\sqrt{8}\approx{2.83}\)
\(\sqrt{10}\approx{3.16}\)
----------------------------------------------------------------Some rules of engagement with exponents are given below:
Laws of Exponents:- \(x^A * x^B = x^{(A+B)}\)
- \(\frac{x^A}{x^B} = x^{(A-B)}\)
- \(x^a * y^a = (xy)^a\)
- \(x^{(-a)} = \frac{1}{x^a}\)
- \(x^0 = 1\)
- \(x^1 = x\)
- \((x^A)^B = x^{(AB)}\)
- \(x^{\frac{a}{b}}\) = \(^b\sqrt{x^a}\)
The above rules are the exhaustive lists of concepts you can use to tackle any exponents problem. However, there are some tricky versions of the above rules which may fool you during your exams which are listed below:
Negative Bases
- \(x^A = +ve\) value; if \(x\) is -ve and \(A\) is even
- \(x^A = -ve\) value; if \(x\) is -ve and \(A\) is odd
For example: \((-3)^2 = 9\), \((-3)^3 = -27\)
Keep in mind \((-3)^2 = 9\), whereas \(-3^2 = -9\).
\(-3^2\) simply means negative of square of 3.
Square Roots
Square root is basically the inverse of a square. In terms of functions if we write,
\(f(x) = x^2\)
Then the inverse of the function will be
\(f^{-1}(x) = \sqrt{x}\)
Algebraic way of writing a square root is simply powering it to the power of \(\frac{1}{2}\), i.e,
\(x^{1/2} = \sqrt{x}\)
However, there are some pitfalls when it comes to square roots, like the following example:
- \(\sqrt{x} = r\); such that \(r^2 = x\), where \(r\) is nonnegetive.
Keep in mind eventhough \((-r)^2 = x\), symbol \(\sqrt{x}\) is used denote the nonegetive root of a number.
For example: \(\sqrt{9} = 3\); not -3.
Some basic properties of square roots are given below:
- \((\sqrt{x})^2 = x\)
- \(\sqrt{x^2} = x\)
- \(\sqrt{x}\sqrt{y} = \sqrt{xy}\)
- \(\frac{\sqrt{x}}{\sqrt{y}}= \sqrt{\frac{x}{y}}\)