Exponents and Logarithms
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The ACT Mathematics test is a timed exam...60 questions in 60 minutes
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the correct answer by eliminating the incorrect options.
Other suggestions are listed in the solutions/explanations as applicable.
These are the solutions to the ACT past questions on the topics: Exponents and Logarithms.
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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)
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Law 1: Exponents
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(1.) Product Rule
$ p^c * p^d = p^{c + d} \\[4ex] p^{c + d} = p^c * p^d $
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Law 1: Logarithms
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(1.)
$ \log_p{c} + \log_p{d} = \log_p{cd} \\[4ex] \log_p{cd} = \log_p{c} + \log_p{d} $
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Law 2: Exponents
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(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} $
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Law 2: Logarithms
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(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} $
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Law 3: Exponents
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(3.)
$ {any\: base}^0 = 1 \\[5ex] p^0 = 1 $
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Law 3: Logarithms
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(3.)
$ \log_{any\: base}{1} = 0 \\[5ex] \log_p{1} = 0 \\[4ex] \ln{1} = \log_e{1} = 0 $
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Law 4: Exponents
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(4.)
$ {any\: base}^1 = any\: base \\[5ex] p^1 = p $
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Law 4: Logarithms
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(4.)
$ \log_{any\: base}{any\: base} = 1 \\[5ex] \log_p{p} = 1 $
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Law 5: Exponents
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(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}} $
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Law 5: Logarithms
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(5.)
$ \log_p{c^d} = d * \log_p{c} \\[4ex] d * \log_p{c} = \log_p{c^d} $
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Law 6: Exponents
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(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}} $
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Law 6: Logarithms
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(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} $
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Law 7: Exponents
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(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}} $
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Law 7: Logarithms
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(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.
$ A.\;\; \dfrac{1}{5} \\[5ex] B.\;\; 2 \\[3ex] C.\;\; 5 \\[3ex] D.\;\; 25 \\[3ex] E.\;\; 5^{25} \\[4ex] $
$ \log_5{25} \\[4ex] = \log_5{5^2} \\[5ex] = 2\log_5{5} ...Law\;5...Log \\[4ex] = 2 \cdot 1 ...Law\;4...Log \\[3ex] = 2 $
$ F.\;\; 3 \\[3ex] G.\;\; 8 \\[3ex] H.\;\; 9 \\[3ex] J.\;\; 52 \\[3ex] K.\;\; 108 \\[3ex] $
$ \log_3{54} - \log_3{2} = \log_2{x} \\[4ex] \log_3{54 \div 2} = \log_2{x} ...Law\;2...Log \\[4ex] \log_3{27} = \log_2{x} \\[4ex] \log_2{x} = \log_3{3^3} \\[5ex] \log_2{x} = 3\log_3{3} ...Law\;5...Log \\[4ex] \log_2{x} = 3(1) ...Law\;4...Log \\[4ex] \log_2{x} = 3 \\[4ex] x = 2^3...\text{Relationship between Exponents and Logarithms} \\[4ex] x = 8 \\[3ex] $
$ F.\;\; \dfrac{1}{2} \\[5ex] G.\;\; 1 \\[3ex] H.\;\; \dfrac{3}{2} \\[5ex] J.\;\; 2 \\[3ex] K.\;\; 3 \\[3ex] $
$ 36^{x - 1} = 6 \\[4ex] 6^{2(x - 1)} = 6^1 \\[4ex] \text{same base; equate exponents} \\[3ex] 2(x - 1) = 1 \\[3ex] x - 1 = \dfrac{1}{2} \\[5ex] x = \dfrac{1}{2} + 1 \\[5ex] x = \dfrac{1}{2} + \dfrac{2}{2} \\[5ex] x = \dfrac{3}{2} $
Which of the following expressions is equivalent to $0.0002^n$
$ F.\;\; 2^{-2n} \\[4ex] G.\;\; 2^{-3n} \\[4ex] H.\;\; 2^n \times 10^{-n} \\[4ex] J.\;\; 2^n \times 10^{-3n} \\[4ex] K.\;\; 2^n \times 10^{-4n} \\[4ex] $
$ 0.0002^n \\[4ex] = \dfrac{0\color{darkblue}{.0002}^n}{\color{darkblue}{10000}^n} \\[6ex] = \dfrac{2^n}{(10^4)^n} \\[6ex] = \dfrac{2^n}{10^{4n}}...Law\;5...Exp \\[6ex] = 2^n \times 10^{-4n} ...Law\;6...Exp $
