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Exponents and Logarithms

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These are the solutions to the NSSCO past questions on Exponents and Logarithms.
The TI-84 Plus CE shall be used for applicable questions.
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These are the notable notes regarding factoring

Factoring Formulas

$ \underline{Difference\;\;of\;\;Two\;\;Squares} \\[3ex] (1.)\;\;x^2 - y^2 = (x + y)(x - y) \\[5ex] \underline{Difference\;\;of\;\;Two\;\;Cubes} \\[3ex] (2.)\;\; x^3 - y^3 = (x - y)(x^2 + xy + y^2) \\[5ex] \underline{Sum\;\;of\;\;Two\;\;Cubes} \\[3ex] (3.)\;\; x^3 + y^3 = (x + y)(x^2 - xy + y^2) $

Laws of Exponents and Laws of Logarithms (Math)

  • Law 1: Exponents

  • (1.) Product Rule

    $ p^c * p^d = p^{c + d} \\[4ex] p^{c + d} = p^c * p^d $

  • Law 1: Logarithms

  • (1.)

    $ \log_p{c} + \log_p{d} = \log_p{cd} \\[4ex] \log_p{cd} = \log_p{c} + \log_p{d} $

  • Law 2: Exponents

  • (2.) Quotient Rule

    $ p^c \div p^d = p^{c - d} \\[4ex] \dfrac{p^c}{p^d} = p^{c - d} \\[4ex] p^{c - d} = p^c \div p^d \\[4ex] p^{c - d} = \dfrac{p^c}{p^d} $

  • Law 2: Logarithms

  • (2.)

    $ \log_p{c} - \log_p{d} = \log_p({c \div d}) \\[4ex] \log_p{c} - \log_p{d} = \log_p{\left(\dfrac{c}{d}\right)} \\[5ex] \log_p({c \div d}) = \log_p{c} - \log_p{d} \\[4ex] \log_p{\left(\dfrac{c}{d}\right)} = \log_p{c} - \log_p{d} $

  • Law 3: Exponents

  • (3.)

    $ {any\: base}^0 = 1 \\[5ex] p^0 = 1 $

  • Law 3: Logarithms

  • (3.)

    $ \log_{any\: base}{1} = 0 \\[5ex] \log_p{1} = 0 \\[4ex] \ln{1} = \log_e{1} = 0 $

  • Law 4: Exponents

  • (4.)

    $ {any\: base}^1 = any\: base \\[5ex] p^1 = p $

  • Law 4: Logarithms

  • (4.)

    $ \log_{any\: base}{any\: base} = 1 \\[5ex] \log_p{p} = 1 $

  • Law 5: Exponents

  • (5.) Expanded Power Rule

    $ (p)^c = (p^1)^c = p^{1 * c} = p^c \\[5ex] \left(\dfrac{p}{q}\right)^c = \dfrac{p^c}{q^c} \\[7ex] (p^c)^d = p^{c * d} \\[5ex] p^{c * d} = (p^c)^d \\[7ex] \left(\dfrac{p^c}{q^d}\right)^e = \dfrac{p^{ce}}{q^{de}} \\[7ex] (pk)^d = p^d * k^d \\[5ex] p^d * k^d = (pk)^d \\[5ex] (p^c k^d)^m = p^{cm} * k^{dm} \\[5ex] p^{cm} * k^{dm} = (p^c)^m * (k^d)^m = (p^c k^d)^m \\[5ex] (p^c)^{\dfrac{d}{e}} = p^{\dfrac{cd}{e}} \\[7ex] p^{\dfrac{cd}{e}} = (p^c)^{\dfrac{d}{e}} $

  • Law 5: Logarithms

  • (5.)

    $ \log_p{c^d} = d * \log_p{c} \\[4ex] d * \log_p{c} = \log_p{c^d} $

  • Law 6: Exponents

  • (6.) Rule of Negative Exponents

    $ p^{-c} = \dfrac{1}{p^c} \\[5ex] \dfrac{1}{p^c} = p^{-c} \\[5ex] p^c = \dfrac{1}{p^{-c}} \\[5ex] \dfrac{1}{p^{-c}} = p^c \\[5ex] p^{-\dfrac{c}{d}} = \dfrac{1}{p^{\dfrac{c}{d}}} \\[7ex] \dfrac{1}{p^{\dfrac{c}{d}}} = p^{-\dfrac{c}{d}} \\[7ex] p^{\dfrac{c}{d}} = \dfrac{1}{p^{-\dfrac{c}{d}}} \\[7ex] \dfrac{1}{p^{-\dfrac{c}{d}}} = p^{\dfrac{c}{d}} $

  • Law 6: Logarithms

  • (6.)
    Change of Base of Log

    $ \log_p{d} = \dfrac{\log_c{d}}{\log_c{p}} \\[4ex] \dfrac{\log_c{d}}{\log_c{p}} = \log_p{d} \\[4ex] \log_p{d} * \log_c{p} = \log_c{d} \\[4ex] \log_c{d} = \log_p{d} * \log_c{p} $

  • Law 7: Exponents

  • (7.) Rule of Fractional Exponents

    $ p^{\dfrac{1}{c}} = \sqrt[c]{p} \\[5ex] p^{\dfrac{c}{d}} = \sqrt[d]{p^c} \\[5ex] p^{\dfrac{c}{d}} = (\sqrt[d]{p})^c \\[5ex] \sqrt[d]{p^c} = p^{\dfrac{c}{d}} \\[5ex] (\sqrt[d]{p})^c = p^{\dfrac{c}{d}} $

  • Law 7: Logarithms

  • (7.)

    $ p^{\log_p{c}} = c \\[5ex] c = p^{\log_p{c}} \\[5ex] p^{d\log_p{c}} = p^{\log_p{c^d}} = c^d \\[5ex] c^d = p^{\log_p{c^d}} = p^{d\log_p{c}} \\[5ex] e^{\ln{c}} = c \\[5ex] c = e^{\ln{c}} \\[5ex] e^{d\ln{c}} = e^{\ln{c^d}} = c^d \\[5ex] c^d = e^{\ln{c^d}} = e^{d\ln{c}} $

Laws of Exponents

Law 1:
If two or more expressions have: the same base, and are being multiplied;
Keep the base
Add the exponents.

Law 2:
If two expressions have: the same base, and are being divided;
Keep the base
Subtract the exponents.
Subtracting the exponents means subtract the exponent of the denominator from the exponent of the numerator.

Law 3:
Any base raised to an exponent of $0$ gives a result of $1$
In other words, the result of any base raised to an exponent of 0, is 1

Law 4:
Any base raised to an exponent of $1$ is that base.
In other words, the result of any base raised to an exponent of 1, is that base.

Law 5:
If an expression, having one base enclosed in parenthesis, has two exponents: an inner exponent and an outer exponent;
Keep the base
Multiply the exponents.
If an expression, having a product of multiple bases enclosed in parenthesis, has two exponents: an inner exponent and an outer exponent;
Keep each base as a product
Multiply the exponents.

Law 6:
A base with a negative exponent is the reciprocal of the same base with the corresponding positive exponent.
A base with a positive exponent is the reciprocal of the same base with the corresponding negative exponent.

Law 7:
Fractional exponents leads to radicals.
In other words, a base whose exponent is a fraction can be expressed as a radical.

Laws of Logarithms

Law 1:
If two logarithms have: the same base, and are being added;
Keep the base
Multiply the numbers.

Law 2:
If two logarithms have: the same base, and are being subtracted;
Keep the base
Divide the numbers.

Law 3:
The logarithm of the number, 1; to any base gives a result of 0
In other words, the result of the logarithm of 1; to any base is 0

Law 4:
The logarithm of a number; to that number as the base, gives a result of 1
In other words, the result of the logarithm of a number to that same number as the base is 1

Law 5:
The logarithm of a number to a base, raised to an exponent; is equal to the exponent times the logarithm of that number to the base.

Law 6:
This Law deals with the Change of Base of Logarithms.
Any logarithm of a number to a base, can be expressed as a ratio of two logarithms to another base.
Say you have the logarithm of a number say $d$ to a base, say p; and
You want to change the base of that logarithm to another base, say c; then the logarithm of d to base p is the logarithm of d to base c divided by the logarithm of p to base c.
In other words, the logarithm of d to base p is the ratio of the logarithm of d to base c, to the logarithm of p to base c.

Law 7:
This Law actually deals with both exponents and logarithms.
The result of a base, raised to an exponent of the logarithm of a number whose base is the base, is the number.

(1.) Solve the following equations.

$ (a.)\;\; 2^x = \dfrac{1}{4} \\[5ex] (b.)\;\; 3^x = 1.5 \\[3ex] $

$ (a.) \\[3ex] 2^x = \dfrac{1}{4} \\[5ex] 2^x = 4^{-1} ...Law\;6...Exp \\[4ex] 2^x = 2^{2(-1)} \\[3ex] 2^x = 2^{-2} ...Law\;5...Exp \\[4ex] \text{same base, equate exponents} \\[3ex] x = -2 \\[3ex] $ Check
$x = -2$
LHS RHS
$ 2^x \\[3ex] 2^{-2} \\[4ex] \dfrac{1}{2^2}...Law\;6...Exp \\[5ex] \dfrac{1}{4} $ $\dfrac{1}{4}$


$ (b.) \\[3ex] 3^x = 1.5 \\[3ex] \text{Introduce Logarithm to both sides} \\[3ex] \log 3^x = \log 1.5 \\[4ex] x \log 3 = \log 1.5 ...Law\;5...Log \\[3ex] x = \dfrac{\log 1.5}{\log 3} \\[5ex] x = 0.3690702464 \\[3ex] x \approx 0.369...\text{to 3 significant figures} \\[3ex] $ Check
$x = 0.3690702464$
LHS RHS
$ 3^x \\[3ex] 3^{0.3690702464} \\[4ex] 1.5 $ $1.5$
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