Exponents and Logarithms

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These are the solutions to Mathematics questions on the topics: Exponents and Logarithms.
When applicable, the TI-84 Plus CE calculator (also applicable to TI-84 Plus calculator) solutions are provided for some 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.) Resuelvan las ecuaciones aplicando previamente cambio de base.

$ \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 $

(2.) If $\log_{10}5 = x$, determine the value of $\log_2{5}$ in terms of x

$ \log_{10} 5 = \dfrac{\log_2{5}}{\log_2 {10}}...Law\;6...Log \\[5ex] ........................................................... \\[3ex] \log_2 {10} \\[3ex] = \log_2{(2 * 5)} \\[3ex] = \log_2{2} + \log_2{5} ...Law\;1...Log \\[3ex] = 1 + \log_2{5} ...Law\;4...Log \\[3ex] ........................................................... \\[3ex] \log_{10} 5 = \dfrac{\log_2{5}}{1 + \log_2{5}} \\[5ex] x = \dfrac{\log_2{5}}{1 + \log_2{5}} \\[5ex] x(1 + \log_2{5}) = \log_2{5} \\[3ex] x + x\log_2{5} = \log_2{5} \\[3ex] x = \log_2{5} - x\log_2{5} \\[3ex] x = \log_2{5}(1 - x) \\[3ex] \log_2{5} = \dfrac{x}{1 - x} $

(3.) In a magic square, the numbers in each row, the numbers in each column, and the numbers on each diagonal have the same sum.
Given the magic square shown below with a, b, c, x, y, z > 0, determine the product xyz in terms of a, b and c

Number 3

$ \text{Row 1 = Row 2} \\[3ex] \log a + \log b + \log x = p + \log y + \log c \\[3ex] p = \log a + \log b + \log x - \log y - \log c ...eqn.(1) \\[5ex] \text{Column 1 = Right Diagonal} \\[3ex] \log a + p + \log z = \log z + \log y + \log x \\[3ex] p = \log y + \log x - \log a ...eqn.(2) \\[5ex] p = p \implies eqn.(1) = eqn.(2) \\[3ex] \log a + \log b + \log x - \log y - \log c = \log y + \log x - \log a \\[3ex] \log a + \log a + \log b - \log c = \log y + \log y \\[3ex] 2\log a + \log b - \log c = 2\log y \\[3ex] \log a^2 + \log b - \log c = \log y^2 ...Law\;5...Log \\[3ex] \log\left(\dfrac{a^2b}{c}\right) = \log y^2 ...Laws\;1\;\;and\;\;2...Log \\[5ex] y^2 = \dfrac{a^2b}{c} ...eqn.(3) \\[5ex] \text{Left Diagonal = Column 3} \\[3ex] \log a + \log y + r = \log x + \log c + r \\[3ex] \log a + \log y = \log x + \log c \\[3ex] \log (ay) = \log (xc) ...Law\;1...Log \\[3ex] ay = xc ...eqn.(4) \\[5ex] \text{Row 3 = Column 3} \\[3ex] \log z + q + r = \log x + \log c + r \\[3ex] q = \log x + \log c - \log z ...eqn.(5) \\[5ex] \text{Column 2 = Row 1} \\[3ex] \log b + \log y + q = \log a + \log b + \log x \\[3ex] q = \log a + \log x - \log y...eqn.(6) \\[5ex] q = q \implies eqn.(5) = eqn.(6) \\[3ex] \log x + \log c - \log z = \log a + \log x - \log y \\[3ex] \log c - \log z = \log a - \log y \\[3ex] \log\left(\dfrac{c}{z}\right) = \log\left(\dfrac{a}{y}\right)...Law\;2...Log \\[5ex] \dfrac{c}{z} = \dfrac{a}{y} \\[5ex] az = cy ...eqn.(7) \\[5ex] eqn.(4) \div eqn.(7) \implies \\[3ex] \dfrac{ay}{az} = \dfrac{xc}{cy} \\[5ex] \dfrac{y}{z} = \dfrac{x}{y} \\[5ex] y^2 = xz ...eqn.(8) \\[5ex] y^2 = y^2 \implies eqn.(3) = eqn.(8) \\[3ex] xz = \dfrac{a^2b}{c}...eqn.(9) \\[5ex] From\;\;eqn.(3) \\[3ex] y^2 = \dfrac{a^2b}{c} ...eqn.(3) \\[5ex] y = \sqrt{\dfrac{a^2b}{c}} \\[5ex] y = \dfrac{a\sqrt{b}}{\sqrt{c}} ...eqn.(10) \\[5ex] \text{From eqns. (9) and (10)} \\[3ex] xz * y = \dfrac{a^2b}{c} * \dfrac{a\sqrt{b}}{\sqrt{c}} \\[5ex] xyz = \dfrac{a^3b\sqrt{b}}{c\sqrt{c}} \\[5ex] xyz = \dfrac{a^3b\sqrt{b}}{c\sqrt{c}} * \dfrac{\sqrt{c}}{\sqrt{c}} \\[5ex] xyz = \dfrac{a^3b\sqrt{bc}}{c^2} $

(4.) Resolva a equação $2^{\log_x{(x^2 - 6x + 9)}} = 3^{2 \cdot \log_x{\sqrt{x}} - 1}$.

$ 2^{\log_x{(x^2 - 6x + 9)}} = 3^{2 \cdot \log_x{\sqrt{x}} - 1} \\[5ex] ................................................................ \\[3ex] \underline{\text{Exponent of the LHS}} \\[3ex] \log_x{(x^2 - 6x + 9)} \\[3ex] = \log_x{(x - 3)(x - 3)} \\[3ex] = \log_x{(x - 3)^2} \\[5ex] \underline{\text{Exponent of the RHS}} \\[3ex] 2 \cdot \log_x{\sqrt{x}} - 1 \\[3ex] = \log_x{(\sqrt{x})^2} - 1 ...Law\;5...Log \\[3ex] = \log_x{x} - 1 \\[3ex] = 1 - 1 ...Law\;4...Log \\[3ex] = 0 \\[5ex] \underline{\text{RHS}} \\[3ex] 3^{2 \cdot \log_x{\sqrt{x}} - 1} \\[3ex] = 3^0 \\[3ex] = 1 ...Law\;3...Exp \\[3ex] ................................................................ \\[3ex] 2^{\log_x{(x - 3)^2}} = 1 \\[3ex] \text{Introduce Logarithms to both sides} \\[3ex] \log [2^{\log_x{(x - 3)^2}}] = \log (1) \\[3ex] \log_x{(x - 3)^2} \cdot \log (2) = \log (1) ...Law\;5...Log \\[3ex] \log_x{(x - 3)^2} = \dfrac{\log(1)}{\log(2)} \\[5ex] \log_x{(x - 3)^2} = \dfrac{0}{\log(2)}...Law\;3...Log \\[5ex] \log_x{(x - 3)^2} = 0 \\[3ex] (x - 3)^2 = x^0 ...\text{Relationship between Exponents and Logarithms} \\[3ex] (x - 3)^2 = 1 ...Law\;3...Exp \\[3ex] x - 3 = \pm \sqrt{1} \\[3ex] x = 3 \pm 1 \\[3ex] x = 3 + 1 \hspace{2em}OR\hspace{2em} x = 3 - 1 \\[3ex] x = 4 \hspace{3em}OR\hspace{3em} x = 2 $

(5.) Solve the equation $\dfrac{A}{B + Ce^{-kt}} = D$ for t in terms of A, B, C, D, and k with natural logarithm.

$ \dfrac{A}{B + Ce^{-kt}} = D \\[5ex] A = D(B + Ce^{-kt}) \\[3ex] A = BD + CDe^{-kt} \\[3ex] CDe^{-kt} = A - BD \\[3ex] e^{-kt} = \dfrac{A - BD}{CD} \\[5ex] \text{Introduce natural logarithm to both sides} \\[3ex] \ln e^{-kt} = \ln\left[\dfrac{A - BD}{CD}\right] \\[5ex] -kt = \ln\left[\dfrac{A - BD}{CD}\right]...Law\;5...Log \\[5ex] t = \dfrac{\ln\left[\dfrac{A - BD}{CD}\right]}{-k} \\[6ex] t = \dfrac{-\ln\left[\dfrac{A - BD}{CD}\right]}{k} \\[5ex] OR \\[3ex] t = \dfrac{-\ln[(A - BD) - \ln(CD)]}{k}...Law\;2...Log \\[5ex] t = \dfrac{-\ln(A - BD) + \ln(CD)}{k} \\[5ex] t = \dfrac{\ln(CD) - \ln(A - BD)}{k} \\[5ex] t = \dfrac{\ln(C) + \ln(D) - \ln(A - BD)}{k}...Law\;1...Log $

(6.) Solve for x if $\log_x\left[\dfrac{\log_4(x)}{\log_4(x) - 3}\right]^{\log_3(x)} = 2$

$ \log_x\left[\dfrac{\log_4(x)}{\log_4(x) - 3}\right]^{\log_3(x)} = 2 \\[5ex] \log_3(x) * \log_x\left[\dfrac{\log_4(x)}{\log_4(x) - 3}\right] = 2 ...Law\;5...Log \\[5ex] \log_3\left[\dfrac{\log_4(x)}{\log_4(x) - 3}\right] = 2 ...Law\;6...Log \\[5ex] \dfrac{\log_4(x)}{\log_4(x) - 3} = 3^2 ...\text{Relationship between Exponents and Logarithms} \\[5ex] \log_4(x) = 9[\log_4(x) - 3] \\[3ex] \log_4(x) = 9\log_4(x) - 27 \\[3ex] 9\log_4(x) - \log_4(x) = 27 \\[3ex] 8\log_4(x) = 27 \\[3ex] \log_4(x) = \dfrac{27}{8} \\[5ex] x = 4^{\dfrac{27}{8}}...\text{Relationship between Exponents and Logarithms} $