Indices, Logarithms, and Surds
Hi guys. Today we’ll be looking to completely exhaust indices, logarithm, and surds based on jamb mathematics syllabus.
If you feel you need a good textbook instead, see the recommended textbook jamb asked students to read for mathematics.
However, for those who feel they have completely exhausted this topic, click here to see tough questions on indices, logarithms, and surds from jamb past question with solution
Having said all that, let’s proceed.
For the best guide in your studies and mathematics tutorials with videos, visit ogschools.com
1.1 Indices
This is the arithmetic operation of raising a number to a power. It is derived from repetitive multiplication. For example: 10\times10\times10\times10=10^4.
The power is also called an index and the number to be raised to the power is called the base.
Here, the number 4 is the power(index) and 10 is the base.
So, 4\times4\times4\times4\times4\times4=4^6 because the number 4(base) is multiplied by itself 6 times(power or index).
Laws of indices
1. a^1=a where a is a real number
2^1=2,\hspace{1pt} 4^1=4,\hspace{1pt} 0^1=0,\hspace{1pt} -10^1=-10
2. a^0=1 where a\neq0
5^0=1, \hspace{2pt} 8^0=1, \hspace{2pt} -20^0=1, \hspace{2pt} 2000^0=1
3. a^x\times a^y=a^{x+y}
4^2\times 4^3=4^{2+3}=4^5
5^0\times 5^1=5^{0+1}=5^1
6^3\times 6^{-1}=6^{3-1}=6^2
4. a^x\div a^y=a^{x-y}
4^2\div 4^3=4^{2-3}=4^{-1}
5^0\div 5^1=5^{0-1}=5^-1
6^3\div 6^{-1}=6^{3-(-1)}=6^{3+1}=6^4
5. a^{-x}=\frac{1}{a^x}
4^{-1}=\frac{1}{4^1}
6^{-4}=\frac{1}{6^4}
5^{-3}=\frac{1}{5^3}
6. (a^x)^y=a^{xy}
(5^2)^3=5^{2\times3}=5^6
(2^1)^{-4}=2^{1\times-4}=2^-4=\frac{1}{2^4}
(5^3)^8=5^{3\times 8}=5^24
7. a^{\frac{1}{x}}=\sqrt[x]{a}
5^{\frac{1}{2}}=\sqrt{5}
2^{\frac{1}{5}}=\sqrt[5]{2}
7^{\frac{1}{6}}=\sqrt[6]{7}
Combining all these rules can lead to other rules like:
a^\frac{x}{y}=\sqrt[y]{a^x}=(\sqrt[y]{a})^x \\ a^\frac{-1}{y}=\frac{1}{a^\frac{1}{y}}=\frac{1}{\sqrt[y]{a}}\\ a^\frac{-x}{y}=\frac{1}{a\frac{x}{y}}=\frac{1}{\sqrt[y]{a^x}}=\frac{1}{(\sqrt[y]{a})^x} \\ (\frac{a}{b})^{-x}=(\frac{b}{a})^x=\frac{b^x}{a^x}\\ (\frac{a}{b})^{\frac{-x}{y}}=(\frac{b}{a})^{\frac{x}{y}}=\frac{(\sqrt[y]{b})^x}{(\sqrt[y]{a})^x}=\sqrt[y]{(\frac{b}{a})^x}=(\sqrt[y]{\frac{b}{a}})^x
Example 2.1
Simplify the following:
1. x^3\div x^{-5}
2. 15\times10^4\div (3\times 1o^{-2})
3. \frac{9\times 10^9}{3\times 10^3}
4. (3m^4)^2
5. (-d^5)^4
6. (-u^3v^2)^4
7. \frac{-(d^2)^3}{d^4\times(-d)}
8. 2^{-2}\times 2^3
9. 2a^{-1}\times(3a)^2
10. \sqrt[3]{4^{1\cdot5}}
11. 125^{1\div3}
12. 0\cdot04^{\frac{1}{2}}
13. (\frac{8}{27})^{\frac{-2}{3}}
14. \sqrt{(125^2)^{\frac{-1}{3}}}
15. (\frac{18}{32})^{\frac{-3}{2}}
Solution
1. x^3\div x^{-5} =x^{3-(-5)}=x^{3+5}=x^8
2. 15\times10^4\div (3\times 10^{-2})= \frac{15\times10^4}{3\times10^{-2}} =\frac{15}{3}\times10^{4-(-2)}=5\times10^{4+2}=5\times10^6
3. \frac{9\times 10^9}{3\times 10^3}=\frac{9}{3}\times10^{9-3}=3\times10^6
4. (3m^4)^2 = 3^2m^{4\times2}=9m^8
5. (-d^5)^4 = d^{5\times4} = d^{20}
6. (-u^3v^2)^4 = (-u^3)^4(v^2)^4 = u^{12}v^8
7. \frac{-(d^2)^3}{d^4\times(-d)} = \frac{-d^{2\times3}}{-d^{4+1}} = \frac{-d^6}{-d^5} = d^{6-5} =d
8. 2^{-2}\times 2^3 = 2^{-2+3} =2
9. 2a^{-1}\times(3a)^2 = 2a^{-1} \times 3^2a^2 = (2\times3^2)a{-1+2} =18a
10. \sqrt[3]{4^{1\cdot5}} = 4^{\frac{1\cdot5}{3}} = 4^{0\cdot5}=4^{\frac{1}{2}}= \sqrt{4}=2
11. 125^{1\div3} = 125^{\frac{1}{3}} = (5^3)^{\frac{1}{3}}=5
12. 0\cdot04^{\frac{1}{2}} = (\frac{4}{100})^{\frac{1}{2}} = (\frac{2}{10})^{2\times\frac{1}{2}} = \frac{2}{10} = \frac{1}{5}
13. (\frac{8}{27})^{\frac{-2}{3}} =(\frac{27}{8})^{\frac{2}{3}} = (\frac{3}{2})^{3\times\frac{2}{3}} = (\frac{3}{2})^2 = \frac{9}{4}
14. \sqrt{(125^2)^{\frac{-1}{3}}} = ((125^2)^{\frac{-1}{3}})^\frac{1}{2} = 125^{2\times\frac{-1}{3}\times\frac{1}{2}} = 125^{\frac{-1}{3}} =(5^3)^{\frac{-1}{3}}= 5^{-1}=\frac{1}{5}
15. (\frac{18}{32})^{-{\frac{3}{2}}} = (\frac{9}{16})^{-{\frac{3}{2}}} = (\frac{16}{9})^{\frac{3}{2}} = (\frac{4}{3})^{2\times\frac{3}{2}} = (\frac{4}{3})^3 = \frac{64}{27}
Example 2.2
Rewrite the following expressions with the positive index only
1. pq^{-2}
2. ab^{-3}
3. (\frac{a}{3b})^{-2}
Solution
1. pq^{-2} =\frac{p}{q^2}
2. ab^{-3} = \frac{a}{b^3}
3. (\frac{a}{3b})^{-2} = (\frac{3b}{a})^2
1.1.1 Standard form
A number is said to be in standard form when it is written as a\times10^n where 0<a<10 and n is an integer. For example 2\times 10^0 and 5.68\times 10^{-4} are numbers in standard form.
Let’s convert some numbers to standard form.
1. 1903000 = 1\cdot903\times 10^6
2. 3\cdot5=3\cdot5\times10^0
3. 0\cdot006313=6\cdot313\times10^-3
4. 307=3\cdot07\times 10^2
1.2 Logarithms
If a=b^c, it is read as a equals b raise to the power of c. It can also be read as c is the logarithm of number a to the base of b.
Note: b=base, c=exponent(power)
\therefore a=b^c\rightarrow c=\log _{b}{a}
We can also say logarithm is the opposite of indices because where 25=5^2, 2=\log_{5}{25}
Laws of logarithm
From our idea on indices, let’s derive some rules in logarithm.
If x=a^b and y=a^c
then b=\log_{a}{x} and c=\log_{a}{y}
1. xy=a^b\times a^c=a^{b+c} \\b+c=\log_{a}{xy}
Similarly, b+c=\log_{a}{x}+\log_{a}{y} \\ \therefore \log_{a}{x}+\log_{a}{y}=\log_{a}{xy}
2. x\div y=\frac{a^b}{a^c}=a^{b-c} \\ b-c=\log_{a}{\frac{x}{y}}
Similarly, b-c= \log_{a}{x}-\log_{a}{y} \\ \therefore \log_{a}{x}-\log_{a}{y}=\log_{a}{\frac{x}{y}}
3. x^n=a^{bn} \\bn=\log_{a}{x^n}
Similarly, bn=n\log_{a}{x} \\ \therefore \log_{a}{x^n}=n\log_{a}{x}
4. \log_{a}{a}=1
5. \log_{a}{a^x}=x\log_{a}{a}
6. If x=0,
\log_{a}{a^0}=0\log_{a}{a}=0
7. a^{\log_{a}{x}}=x
8. If a^c=b, then a=b^{\frac{1}{c}} \\ \rightarrow c=\log_{a}{b}, and \frac{1}{c}=\log_{b}{a} \\ \therefore \frac{1}{\log_{a}{b}}=\log_{b}{a} and \frac{1}{\log_{b}{a}}=\log_{a}{b}
9. Change of base
let a=b^c, then c=\log_{b}{a}
Similarly, let x=a^d, then d=\log_{a}{x} \\ \therefore x=a^d=(bc)^d=b^{cd} \\cd=\log_{b}{x}
Similarly, cd= \log_{b}{a}\times \log_{a}{x} \\ \rightarrow \log_{b}{a}\times \log_{a}{x}=\log_{b}{x}\\ \therefore \log_{a}{x}=\frac{\log_{b}{x}}{\log_{b}{a}}
We just changed the logarithm of a number from \log_{a} to \log_{b}
10. \log_{a^y}{x}=\frac{\log_{a}{x}}{\log_{a}{a^y}}=\frac{\log_{a}{x}}{y\log_{a}{a}}=\frac{1}{y}\log_{a}{x}
11. \log_{a^{-y}}{x}=\frac{\log_{a}{x}}{\log_{a}{a^{-y}}}=\frac{\log_{a}{x}}{-y\log_{a}{a}}=-\frac{1}{y}\log_{a}{x}
Example 2.3
1. Simplify \log_{a}{b}+\log_{a}{bx}+\log_{a}{bx^2}
2. Simplify as a logarithm in base 3: \log_{\frac{1}{3}}{7} +2\log_{9}{49}-\log_{\sqrt{3}}{\frac{1}{7}}
3. Evaluate
a. 81^{\frac{1}{\log_{2}{3}}}\div 36^{\frac{1}{\log_{2}{6}}}
b. \log_{4}{\frac{1}{8}}
c. \log_{\frac{1}{3}}{27}
4. \frac{1}{\log_{a}{ab}}+\frac{1}{\log_{b}{ab}}
Solution
Go through all the laws again for easier understanding. Now, let’s proceed
1. \log_{a}{b}+\log_{a}{bx}+\log_{a}{bx^2}\\ \log_{a}{(b\times bx\times bx^2)}// \log_{a}{b^3x^3}\\ \log{a}{(bx)^3} \\ 3log_{a}{bx}
2. \log_{\frac{1}{3}}{7} +2\log_{9}{49}-\log_{\sqrt{3}}{\frac{1}{7}}\\ \log_{3^{-1}}{7} +2\log_{3^2}{49}-\log_{3^{\frac{1}{2}}}{\frac{1}{7}}\\ -\log_{3}{7} +\frac{2}{2}\log_{3}{49}-\frac{1}{\frac{1}{2}}\log_{3}{\frac{1}{7}}\\ -\log_{3}{7} +\log_{3}{49}-2\log_{3}{\frac{1}{7}}\\ \log_{3}{49}-\log_{3}{7}-2\log_{3}{\frac{1}{7}}\\ \log_{3}{49}-\log_{3}{7}-\log_{3}{(\frac{1}{7})^2}\\ \log_{3}{49}-\log_{3}{7}-\log_{3}{\frac{1}{49}}\\ \log_{3}{(49\div 7\div \frac{1}{49})}\\ \log_{3}{(49\times \frac{1}{7}\times 49)}\\ \log_{3}{7\times49}\\ \log_{3}{7^3}\\ 3\log_{3}{7}
3.
a. 81^{\frac{1}{\log_{2}{3}}}\div 36^{\frac{1}{\log_{2}{6}}}\\ 81^{\log_{3}{2}}\div 36^{\log_{2}{6}} \\ 3^{4\log_{3}{2}}\div 6^{2\log_{2}{6}} \\ 3^{\log_{3}{2^4}}\div 6^{\log_{2}{6^2}} \\ 2^4\div 2^2 \\ 2^{4-2}=2^2=4
b. \log_{4}{\frac{1}{8}}\\ \log_{2^2}{8^{-1}}\\ \log_{2^2}{2^{-3}}\\ -\frac{3}{2}\log_{2}{2}\\ -\frac{3}{2}
c. \log_{\frac{1}{3}}{27}\\ \log_{3^{-1}}{3^3} \\ -3\log_{3}{3} \\ -3
4. \frac{1}{\log_{a}{ab}}+\frac{1}{\log_{b}{ab}}\\ \log_{ab}{a}+\log_{ab}{b}\\ \log_{ab}{ab} \\ 1
Solving equations involving indices and logarithms
Example 2.4
Find the unknown in the following equations
1. x^{\frac{1}{2}}=2
2. 5x=40x^{-\frac{1}{2}}
3. 4^{x-1}=64
4. 9^x=27
5. 2\log_{a}{x}-\log_{a}{x-1}=\log_{a}{x-2}
6. 4\log_{a}{\sqrt{x}}-\log_{a}{3x} =\log_{a}{x^{-2}}
7. Find y in terms of x: 2\log_{a}{y}-3\log_{a}{x^2}=\log_{a}{\sqrt{y}}+\log_{a}{x}
8. For what range of x is \log_{a}{x-3} valid?
Solution
1. x^{\frac{1}{2}}=2
Squaring both sides
(x^{\frac{1}{2}})^2=2^2\\ \therefore x=4
2. 5x=40x^{-\frac{1}{2}} \\ \frac{x}{x^{-\frac{1}{2}}}=\frac{40}{5} \\ x^{1+\frac{1}{2}}=8 \\ x^{\frac{3}{2}}=8 \\ x=8^{\frac{2}{3}} \\ x=4
3. 4^{x-1}=64 \\ 4^{x-1}=4^3 \\ x-1=3 \\ x=4
4. 9^x=27 \\ 3^{2x}=3^3 \\2x=3 \\ x=\frac{3}{2}
5. 2\log_{a}{x}-\log_{a}{x-1}=\log_{a}{x-2} \\ \log_{a}{x^2}=\log_{a}{x-2}+\log_{a}{x-1} \\ \log_{a}{x^2}=\log_{a}{(x-2)(x-1)} \\ \log_{a}{x^2}=\log_{a}{(x^2-3x+2)} \\ x^2=x^2-3x+2 \\ 3x-2=0 \\ 3x=2 \\ x=\frac{2}{3}
6. 4\log_{a}{\sqrt{x}}-\log_{a}{3x} =\log_{a}{x^{-2}} \\ \log_{a}{(\sqrt{x})^4}-\log_{a}{3x} =\log_{a}{x^{-2}} \\ \log_{a}{x^2}=\log_{a}{x^{-2}}+\log_{a}{3x} \\ \log_{a}{x^2}=\log_{a}{x^{-2}}+\log_{a}{3x} \\ \log_{a}{x^2}=\log_{a}{\frac{3x}{x^2}} \\ x^2=\frac{3x}{x^2} \\ x^2=\frac{3}{x} \\ x^3=3 \\ x=\sqrt[3]{3}
7. 2\log_{a}{y}-3\log_{a}{x^2}=\log_{a}{\sqrt{y}}+\log_{a}{x} \\ \log_{a}{y^2}-\log_{a}{x^6}=\log_{a}{\sqrt{y}}+\log_{a}{x} \\ \log_{a}{\frac{y^2}{x^{-6}}}=\log_{a}{x\sqrt{y}} \\ \frac{y^2}{x^6}=x\sqrt{y} \\ \frac{y^2}{\sqrt{y}}=x^7 \\ y^{\frac{3}{2}}=x^7 \\ y^3=x^14 \\ y=\sqrt[3]{x^{14}}
8. You can only find the logarithm of a positive number. For x-3 to be positive, x-3>0
for \log_{a}{x-3} to be valid,
x-3>0 \\ \therefore x>3
1.3 Surds
A general surd is an irrational number of the form a\sqrt[n]{b}, where a is a rational number and \sqrt[n]{b} is an irrational number. \sqrt[n]{} is called the RADICAL.
\sqrt{b} is taken to mean the positive square root of b, the negative square root is -\sqrt{b}.
Examples of surds are \sqrt{5}, \sqrt[6]{8}, 5\sqrt{2}, 8\sqrt[3]{3} etc.
Rules for manipulating surd
1. a\sqrt{b}+c\sqrt{d}=(a+c)\sqrt{b}
2. a\sqrt{b}-c\sqrt{d}=(a-c)\sqrt{b}
3. \sqrt{ab}=\sqrt{a}\times \sqrt{b}
4. a\sqrt{b}\times c\sqrt{d}=ac\sqrt{bd}
5. \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}
Rationalization of a surd
When a number is expressed in the form \frac{a}{\sqrt{b}}, it is difficult to simplify. Since surds are irrational numbers, we rationalize the denominator to give \frac{a\sqrt{b}}{b}, i.e \frac{a}{b}\times \frac{\sqrt{b}}{\sqrt{b}}=\frac{a\sqrt{b}}{b}.
If we have \frac{1}{\sqrt{2}}, it is easier to solve \frac{\sqrt{2}}{2} than to solve \frac{1}{\sqrt{2}} because \frac{\sqrt{2}}{2}=\frac{1.414}{2} and \frac{1}{\sqrt{2}}=\frac{1}{1.414}
Useful hint on rationalization
1. \sqrt{a}\times \sqrt{a}=a
2. (\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b
3. (x\sqrt{a}+y\sqrt{b})(x\sqrt{a}-y\sqrt{b})=x^2a-y^2b
4. (x+y\sqrt{b})(x-y\sqrt{b})=x^2-y^2b
Example 2.5
1. \frac{8-3\sqrt{6}}{2\sqrt{3}+3\sqrt{2}} in the form m\sqrt{3} +n\sqrt{2} where m and n are rational numbers.
2. \frac{2}{(3\sqrt{5}-4)^2} in the form a+b\sqrt{c}, where a,b are rational numbers.
3. Simplify 4\sqrt{\frac{1}{32}}-2\sqrt{\frac{1}{8}}
4. Solve for x if \sqrt{3x+1}-\sqrt{x+4}=1
Solution
1. \frac{8-3\sqrt{6}}{2\sqrt{3}+3\sqrt{2}}= \frac{8-3\sqrt{6}}{2\sqrt{3}+3\sqrt{2}}\times \frac{2\sqrt{3}-3\sqrt{2}}{2\sqrt{3}-3\sqrt{2}}\\ \frac{(8-3\sqrt{6})(2\sqrt{3}-3\sqrt{2})}{(2\sqrt{3}+3\sqrt{2})(2\sqrt{3}-3\sqrt{2})}\\ \frac{16\sqrt{3}-24\sqrt{2}-6\sqrt{18}+9\sqrt{12}}{12-18}\\ \frac{16\sqrt{3}-24\sqrt{2}-6\sqrt{9\times2}+9\sqrt{4\times3}}{-6}\\ \frac{16\sqrt{3}-24\sqrt{2}-18\sqrt{2}+18\sqrt{3}}{-6}\\ \frac{34\sqrt{3}-42\sqrt{2}}{-6}\\ -\frac{17}{3}\sqrt{3}+7\sqrt{2}
2. \frac{2}{(3\sqrt{5}-4)^2}=\frac{2}{(3\sqrt{5}-4)(3\sqrt{5}-4)}\\ \frac{2}{9(5)-12\sqrt{5}-12\sqrt{5}+16}\\ \frac{2}{45+16-24\sqrt{5}}\\ \frac{2}{61-24\sqrt{5}}=\frac{2}{61-24\sqrt{5}}\times\frac{61+24\sqrt{5}}{61+24\sqrt{5}} \\ \frac{122+48\sqrt{5}}{3721-2880}\\ \frac{122+48\sqrt{5}}{841}\\ \frac{122}{841}+\frac{48}{841}\sqrt{5}
3. \sqrt{3x+1}-\sqrt{x+4}=1\\ \sqrt{3x+1}=1+\sqrt{x+4}
Squaring both sides
(\sqrt{3x+1})^2=(1+\sqrt{x+4})^2\\ 3x+1=1+2\sqrt{x+4}+x+4\\ 2x-4=2\sqrt{x+4}\\ x-2=\sqrt{x+4}
Squaring both sides
(x-2)^2=(\sqrt{x+4})^2\\ x^2-4x+4=x+4\\ x^2-4x+4=x+4\\x^2-5x=0\\x(x-5)=0\\x=0, x=5
put x=0 in the given equation
\sqrt{0+1}-\sqrt{0+4}=\sqrt{1}-\sqrt{4}=1-2=-1Similarly, put x=5 in the given equation
\sqrt{15+1}-\sqrt{5+4}=\sqrt{16}-\sqrt{9}=4-3=1\\ \therefore x=5 is the correct answer.
Let’s take a look at why only one of the solutions is correct.
Let x=5, it means x is 5 and nothing else
on squaring both sides,
Squaring both sides made x have two possible answers which are \pm5
But from the initial assumption, x was equal to 5 only.
Same rule applied here which gave us two solutions. However, only one of the solution gave the correct answer.
Square root of a surd
Let’s take a look on how to find the square root of a surd. It comes in two forms
1. a+c\sqrt{d}
Let \sqrt{a+c\sqrt{d}}=\sqrt{x}+\sqrt{y}
Squaring both sides
(\sqrt{a+c\sqrt{d}})^2=(\sqrt{x}+\sqrt{y})^2\\ a+c\sqrt{d}=x+y+2\sqrt{xy}
Comparing both sides of the equation, we have;
a=x+y\\ d=xy\\ c=2
2. a-c\sqrt{d}
Let \sqrt{a-c\sqrt{d}}=\sqrt{x}-\sqrt{y}
Squaring both sides
(\sqrt{a-c\sqrt{d}})^2=(\sqrt{x}-\sqrt{y})^2\\ a-c\sqrt{d}=x+y-2\sqrt{xy}
Comparing both sides of the equation, we have;
a=x+y\\ d=xy\\ c=2
Example 2.6
1. Find the square root of 5+2\sqrt{6}
2. Find the square root of 9-4\sqrt{2}
Solution
1. \sqrt{5+2\sqrt{6}}=\sqrt{x}+\sqrt{y}
Squaring both sides
(\sqrt{5+2\sqrt{6}})^2=(\sqrt{x}+\sqrt{y})^2\\ 5+2\sqrt{6}=x+y+2\sqrt{xy}Comparing both sides of the equation
x+y=5———-eq(1)
xy=6———–eq(2)
From eq(1), x=5-y
Substitute x=5-y into eq(2)
(5-y)y=6\\ 5y-y^2=6\\y^2-5y+6=o\\y^2-2y-3y+6=0\\(y^2-2y)-(3y-6)=0\\y(y-2)-3(y-2)=0\\y-2=0, y-3=0\\y=2, y=3
if y=2, then x=3
if y=3, then x=2\\ \therefore \sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}
2. Let \sqrt{9-4\sqrt{2}}=\sqrt{x}-\sqrt{y}
Squaring both sides
(\sqrt{9-4\sqrt{2}})^2=(\sqrt{x}-\sqrt{y})^2\\ 9-4\sqrt{2}=x+y-2\sqrt{xy}Comparing both sides of the equation
x+y=9———-eq(1)
2\sqrt{xy}=4\sqrt{2}\\ \sqrt{xy}=2\sqrt{2}\\ (\sqrt{xy})^2=(2\sqrt{2})^2\\ xy=8———–eq(2)
From eq(1), x=9-y
put x=9-y into eq(2)
(9-y)y=8\\9y-y^2=8\\y^2-9y+8=0\\y^2-8y-y+8=0\\(y^2-8y)-(y-8)=0\\y(y-8)-1(y-8)=0\\(y-1)(y-8)=0\\y=1, y=8
if y=1, x=8
if y=8, x=1\\ \therefore \sqrt{9-4\sqrt{2}}=\sqrt{8}-\sqrt{1} because it cannot be negative.
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