Exponents and Logarithms
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These are the solutions to Mathematics 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.)\;\; \log_{\dfrac{1}{2}}x + \log_4 x^3 = \log_3 27 \\[5ex] $
$ (a.) \\[3ex] \log_{\dfrac{1}{2}}x + \log_4 x^3 = \log_3 27 \\[5ex] \text{Let us handle one term at a time} \\[3ex] =================================== \\[3ex] \log_{\dfrac{1}{2}}x ...\text{change base of Log two} \\[3ex] \log_{\dfrac{1}{2}}x = \dfrac{\log_2 x}{\log_2 \dfrac{1}{2}}...Law\;6...Log \\[5ex] ............................................................... \\[3ex] \log_2 \dfrac{1}{2} \\[5ex] = \log_2 2^{-1} ...Law\;6...Exp \\[4ex] = -1\log_2 2 ...Law\;5...Log \\[3ex] = -1 \cdot 1 ...Law\;4...Log \\[3ex] = -1 \\[3ex] ................................................................ \\[3ex] \log_{\dfrac{1}{2}}x = \dfrac{\log_2 x}{-1} \\[5ex] = -\log_2 x \\[3ex] =================================== \\[3ex] \log_4 x^3 \\[4ex] = 3\log_4 x \\[3ex] \log_4 x ...\text{change base of Log to two} \\[3ex] \log_4 x = \dfrac{\log_2 x}{\log_2 4}...Law\;6...Log \\[5ex] ................................................................ \\[3ex] \log_2 4 \\[3ex] = \log_2 2^2 \\[4ex] = 2\log_2 2 ...Law\;5...Log \\[3ex] = 2 \cdot 1 ...Law\;4...Log \\[3ex] = 2 \\[3ex] ................................................................ \\[3ex] \log_4 x = \dfrac{\log_2 x}{2} \\[5ex] 3\log_4 x = 3 \cdot \dfrac{\log_2 x}{2} = \dfrac{3\log_2 x}{2} \\[5ex] =================================== \\[3ex] \log_3 27 \\[3ex] = \log_3 3^3 \\[4ex] = 3\log_3 3 ...Law\;5...Log \\[3ex] = 3 \cdot 1 ...Law\;4...Log \\[3ex] = 3 \\[3ex] =================================== \\[3ex] \implies \\[3ex] -\log_2 x + \dfrac{3\log_2 x}{2} = 3 \\[4ex] Substitute\;\;\log_2 x = p \\[3ex] \implies \\[3ex] -p + \dfrac{3p}{2} = 3 \\[5ex] \text{Multiply each term by 2} \\[3ex] -2p + 3p = 6 \\[3ex] p = 6 \\[3ex] But\;\; log_x = p \\[3ex] log_2 x = 6 \\[3ex] x = 2^6 \\[3ex] x = 64 $
