Algebra (Part 5)

November 27, 2025

Zeros of Polynomials

The zeros of a polynomial are the values of the variable (often \(x\)) that make the polynomial's value equal to zero. They are also known as the roots of the polynomial. Graphically, the zeros correspond to the points where the polynomial's graph intersects the x-axis.

Examples

• 

  Zero	Factor
  \((x = 0)\)	\(x\)
  \((x = -\frac{5}{2})\)	\((2x + 5)\)
  \((x = 4)\)	\((x - 4)\)

• 

  \begin{aligned} f(x) &= 2x^{3}-x^{2}-8x+4 \\\\ &= x^{2}(2x-1)-4(2x-1) \\\\ &= (x^{2}-4)(2x-1) \\\\ &= (x+2)(x-2)(2x-1) \end{aligned}

  Zero	Factor
  \((x = -2)\)	\((x + 2)\)
  \((x = 2)\)	\((x - 2)\)
  \((x = \frac{1}{2})\)	\((2x - 1)\)

Positive and Negative Intervals of Polynomials

Positive and negative intervals of a polynomial are the ranges of x-values where the function's graph is above the x-axis (positive) or below the x-axis (negative). To find these intervals, first identify the polynomial's zeros (where it crosses the x-axis) by setting the function to zero and solving for \(x\). Since polynomials are continuous, the sign of the function (positive or negative) is constant between any two consecutive zeros. Test a value within each interval to determine the sign for that entire interval.

Examples

Multiplicity of Zeros of Polynomials

The multiplicity of a zero of a polynomial is the number of times its corresponding factor appears in the polynomial's factored form, which is also its exponent. For example, in the polynomial \(p(x)=(x-3)^{2}(x+1)\), the zero \(x=3\) has a multiplicity of 2 because the factor \((x-3)\) is squared, while the zero \(x=-1\) has a multiplicity of 1. This multiplicity determines how the graph of the polynomial behaves at the x-axis: an even multiplicity causes the graph to touch and turn back ("bounce") at the zero, while an odd multiplicity causes the graph to cross the x-axis.

Examples

There are three lines on that graph.

• Red line, \(f_{1}\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\)
• Green line, \(f_{2}\left(x\right)=\left(x-1\right)\left(x-3\right)^{2}\)
• Purple line, \(f_{3}\left(x\right)=\left(x-1\right)\left(x-3\right)\)

As we can see, the green line has a bounce at \(x = 3\), which is because it has a multiplicity of \(2\) (even) from the factor of \((x - 3)^2\).

End Behavior of Polynomials

The end behavior of a polynomial is determined by its degree (even or odd) and the sign of its leading coefficient (positive or negative). An even degree results in both ends of the graph pointing in the same direction, while an odd degree results in the ends pointing in opposite directions. The leading coefficient determines the direction for the right side (positive leads to up, negative leads to down), and the degree determines the left side's direction relative to the right.

End Behavior of Monomials: \(f(x) = ax^n\)

\(n\) is even and \((a \gt 0)\)	\(n\) is even and \((a \lt 0)\)
as \(x\rightarrow -\infty \), \(f(x)\rightarrow +\infty \) and as \(x\rightarrow +\infty \), \(f(x)\rightarrow +\infty \)	as \(x\rightarrow -\infty \), \(f(x)\rightarrow -\infty \) and as \(x\rightarrow +\infty \), \(f(x)\rightarrow -\infty \)
\(n\) is odd and \((a \gt 0)\)	\(n\) is odd and \((a \lt 0)\)
as \(x\rightarrow -\infty \), \(f(x)\rightarrow +\infty \) and as \(x\rightarrow +\infty \), \(f(x)\rightarrow +\infty \)	as \(x\rightarrow -\infty \), \(f(x)\rightarrow -\infty \) and as \(x\rightarrow +\infty \), \(f(x)\rightarrow -\infty \)

Rational Exponents

\[a^{\frac{1}{n}} = \sqrt[n]{a}\]

\[a^{\frac{m}{n}} = (\sqrt[n]{a})^{m} = \sqrt[n]{a^{m}}\]

Rational exponents are fractional exponents, written as \(p/q\), where the denominator (\(q\)) indicates the root and the numerator (\(p\)) indicates the power. These exponents are equivalent to radicals and can be manipulated using the same rules as integer exponents.

Here are some examples.

• \(\sqrt[4]{x} = x^{\frac{1}{4}}\)
• \(t^{\frac{1}{3}} = \sqrt[3]{t}\)
• \(\sqrt[7]{b} = b^{\frac{1}{7}}\)
• \(x^{\frac{1}{8}} = \sqrt[8]{x}\)
• \(z^{0.2} = \sqrt[5]{z}\)
• \(\sqrt[7]{v^{3}} = (v^{3})^{\frac{1}{7}} = v^{\frac{3}{7}}\)
• \((\sqrt[4]{d})^{3} = \sqrt[4]{d^3} = (d^{\frac{1}{4}})^{3}\)
• \((k^{\frac{1}{8}})^{-1} = (k^{-1})^\frac{1}{8} = (\sqrt[8]{k})^{-1} = k^{-\frac{1}{8}}\)
• \(x^{1.4} = \sqrt[5]{x^{7}} = (\sqrt[5]{x})^{7} = (x^{\frac{1}{5}})^{7}\)

Rewriting Quotient of Powers

Let's rewrite (simplify) the quotient of powers (rational exponents) by using the General Rules of Exponents that we learned in Algebra (Part 2). Here are some examples.

• \(\frac{m^{\frac{7}{9}}}{m^{\frac{1}{3}}} = \frac{m^{\frac{7}{9}}}{m^{\frac{3}{9}}} = m^{\frac{7}{9} - \frac{3}{9}} = m^{\frac{4}{9}}\)
• \(\frac{1}{a^{-\frac{5}{6}}} = a^{-(-\frac{5}{6})} = a^{\frac{5}{6}}\)
• \((z^{-\frac{4}{3}})^{-\frac{6}{5}} = z^{-\frac{4}{3} \cdot -\frac{6}{5}} = z^{\frac{24}{15}}\)

Rewriting Mixed Radical and Exponential Expressions

Let's rewrite (simplify) mixed radicals and exponential expressions by using the General Rules of Radicals that we learned in Algebra (Part 2). Here is an example.

\begin{aligned} (r^{\frac{2}{3}} s^{3})^{2} \sqrt{20r^{4}s^{5}} &= (r^{\frac{2}{3}})^{2} \cdot (s^{3})^{2} \cdot (4 \cdot 5 \cdot r^{4} \cdot s^{4} \cdot s)^{\frac{1}{2}} \\\\ &= r^{\frac{4}{3}} \cdot s^{6} \cdot 4^{\frac{1}{2}} \cdot 5^{\frac{1}{2}} \cdot (r^{4})^{\frac{1}{2}} \cdot (s^{4})^{\frac{1}{2}} \cdot s^{\frac{1}{2}} \\\\ &= r^{\frac{4}{3}} \cdot s^{6} \cdot 2 \cdot \sqrt{5} \cdot r^{2} \cdot s^{2} \cdot \sqrt{s} \\\\ &= 2 \cdot s^{8} \cdot r^{3\frac{1}{3}} \cdot \sqrt{5} \cdot \sqrt{s} \\\\ &= (2 \cdot s^{8.5} \cdot r^{3\frac{1}{3}} \cdot \sqrt{5}) \text{ or } (2 \cdot s^{8} \cdot r^{3} \cdot \sqrt[3]{r} \cdot \sqrt{5s}) \end{aligned}

Evaluating Exponents and Radicals

We just learned about Rational Exponents, let's evaluate them.

• \(64^{\frac{1}{3}} = \sqrt[3]{64} = 4\)
• \(64^{\frac{2}{3}} = (64^{\frac{1}{3}})^{2} = 4^{2} = 16\)
• \((\frac{8}{27})^{-\frac{2}{3}} = (\frac{27}{8})^{\frac{2}{3}} = \frac{27^{\frac{2}{3}}}{8^{\frac{2}{3}}} = \frac{(27^{\frac{1}{3}})^{2}}{(8^{\frac{1}{3}})^{2}} = \frac{3^{2}}{2^{2}} = \frac{9}{4}\)
• \(9^{\frac{1}{2}} = 3\)
• \(9^{-\frac{1}{2}} = \frac{1}{9\frac{1}{2}} = \frac{1}{3}\)
• \((-27)^{-\frac{1}{3}} = \frac{1}{(-27)^{\frac{1}{3}}} = \frac{1}{-3} = -\frac{1}{3}\)
• \(\frac{5}{3}^{2} = \frac{25}{9}\)
• \((\frac{25}{9})^{\frac{1}{2}} = \sqrt{\frac{25}{9}} = \frac{5}{3}\)
• \((\frac{81}{256})^{-\frac{1}{4}} = (\frac{256}{81})^{\frac{1}{4}} = \frac{256^{\frac{1}{4}}}{81^{\frac{1}{4}}} = \frac{4}{3}\)
• \(\frac{256^{\frac{4}{7}}}{2^{\frac{4}{7}}} = (\frac{256}{2})^{\frac{4}{7}} = 128^{\frac{4}{7}} = (128^{\frac{1}{7}})^{4} = 2^{4} = 16\)
• \begin{aligned} 6^{\frac{1}{2}} \cdot (\sqrt[5]{6})^3 &= 6^{\frac{1}{2}} \cdot (6^{\frac{1}{5}})^{3} \\\\ &= 6^{\frac{1}{2}} \cdot 6^{\frac{3}{5}} \\\\ &= 6^{\frac{1}{2} + \frac{3}{5}} \\\\ &= 6^{\frac{5}{10} + \frac{6}{10}} \\\\ &= 6^{\frac{11}{10}} \end{aligned}

Rewriting Exponential Expressions

As \(A \cdot B^{t}\)

The expression \(A\cdot B^{t}\) is a multiplication of a value or matrix \(A\) by a second value or matrix \(B\) raised to the power of \(t\). The variable \(t\) is the exponent, and the expression can be interpreted in different ways depending on the context, most commonly as a single value being multiplied or a matrix multiplication.

To rewrite an exponential expression as \(A\cdot B^{t}\), you use exponent rules to isolate the variable and constant parts. First, group any constant terms together and move them to the front of the expression to form the value of \(A\). Then, use the General Rules of Exponents that we learned in Algebra (Part 2), like the product of powers (\(a^m \cdot a^n = a^{m+n}\)) rule or the power of a power (\((a^m)^n = a^{mn}\)) rule to simplify the remaining parts until you have a single base raised to the power of \(t\), which will be your \(B\) value.

Here is an example.

\begin{aligned} 10 \cdot 9^{\frac{t}{2} + 2} \cdot 5^{3t} &= 10 \cdot 9^{\frac{t}{2}} \cdot 9^{2} \cdot (5^{3})^{t} \\\\ &= 10 \cdot 3^{t} \cdot 81 \cdot 125^{t} \\\\ &= 810 \cdot (3 \cdot 125)^{t} \\\\ &= 810 \cdot 375^{t} \end{aligned}

As \(A \cdot B^{\frac{t}{10} - 1}\)

Here is an example.

\begin{aligned} \frac{1}{32} \cdot 2^{t} &= \frac{1}{32} \cdot 2^{t \cdot \frac{10}{10}} \\\\ &= \frac{1}{32} \cdot 2^{10 \cdot \frac{t}{10}} \\\\ &= \frac{1}{32} \cdot 1024^{\frac{t}{10}} \\\\ &= \frac{1}{32} \cdot 1024^{\frac{t}{10} - 1 + 1} \\\\ &= \frac{1}{32} \cdot 1024^{\frac{t}{10} - 1} \cdot 1024^{1} \\\\ &= \frac{1024}{32} \cdot 1024^{\frac{t}{10} - 1} \\\\ &= 32 \cdot 1024^{\frac{t}{10} - 1} \end{aligned}

Solving Exponential Equations using Exponent Properties (General Rules of Exponents)

Here are examples.

• \begin{aligned} 26^{9x + 5} &= 1 \text{ or } 26^{0} = 1 \\\\ 26^{9x + 5} &= 1^{0} \\\\ 9x + 5 &= 0 \\\\ 9x &= -5 \\\\ x &= -\frac{5}{9} \end{aligned}
• \begin{aligned} 2^{3x + 5} &= 64^{x - 7} \\\\ 2^{3x + 5} &= (2^{6})^{x - 7} \\\\ 2^{3x + 5} &= 2^{6x - 42} \\\\ 3x + 5 &= 6x - 42 \\\\ 3x - 6x &= -42 - 5 \\\\ -3x &= -47 \\\\ x &= \frac{-47}{-3} \\\\ x &= \frac{47}{3} \end{aligned}
• \begin{aligned} 32^{\frac{x}{3}} &= 8^{x - 12} \\\\ (2^{5})^{\frac{x}{3}} &= (2^{3})^{x - 12} \\\\ 2^{\frac{5x}{3}} &= 2^{3x - 36} \\\\ \frac{5x}{3} &= 3x - 36 \\\\ 5x &= 9x - 108 \\\\ 5x - 9x &= -108 \\\\ -4x &= -108 \\\\ x &= \frac{-108}{-4} \\\\ x &= 27 \end{aligned}
• \begin{aligned} \frac{5^{4x + 3}}{25^{9 - x}} &= 5^{2x + 5} \\\\ \frac{5^{4x + 3}}{(5^{2})^{9 - x}} &= 5^{2x + 5} \\\\ \frac{5^{4x + 3}}{5^{18 - 2x}} &= 5^{2x + 5} \\\\ 5^{4x + 3 - (18 - 2x)} &= 5^{2x + 5} \\\\ 4x + 3 - 18 + 2x &= 2x + 5 \\\\ 6x - 15 &= 2x + 5 \\\\ 6x - 2x &= 5 + 15 \\\\ 4x &= 20 \\\\ x &= \frac{20}{4} \\\\ x &= 5 \end{aligned}

Exponential Models

An exponential model describes a quantity that changes by a constant factor over equal intervals of time, representing either growth or decay. These models are used to represent phenomena like population growth, radioactive decay, and compound interest.

Interpreting the Rate of Change

Here are examples.

• Nora uploaded a funny video on her website, which rapidly gains views over time.

  The relationship between the elapsed time, ‍\(t\), in days, since Nora uploaded the video, and the total number of views, ‍\(V(t)\), is modeled by the following function:

  \(V(t) = 100 \cdot (1.53)^{t}\)

  Every day, the number of views grows by a factor of 1.53.

  Here is why.

  The exponential function modeling the number of views is of the form \(V(t) = A \cdot B^{t}\). Therefore, \(A\) ‍determines the initial number of views (when the video was uploaded) and ‍\(B\) determines the daily change in the number of views.

• The chemical element einsteinium-‍\(253\) naturally loses its mass over time. A sample of einsteinium-‍\(253\) had an initial mass of \(540\) ‍grams when we measured it.

  The relationship between the elapsed time ‍\(t\), in months, and the mass, ‍\(M(t)\), in grams, left in the sample is modeled by the following function:

  \(M(t) = 540 \cdot (\frac{1}{8})^{\frac{t}{2.05}}\)

  The sample loses \(\frac{7}{8}\) of its mass every 2.05 months.

  Here is why.

  The modeling function is of the form \(M(t) = A \cdot B^{f(t)}\), where \(B = \frac{1}{8}\) and \(f(t) = \frac{t}{2.05}\).

  Note that each time \(f(t)\) increases by ‍1, the quantity is multiplied by \(B = \frac{1}{8}\).

  Also note that being multiplied by \(\frac{1}{8}\) is the same as losing \(\frac{7}{8}\).

  \(f\) is a linear function whose slope is \(\frac{1}{2.05}\).

  This means that whenever \(t\) ‍increases by ‍\(\Delta t\), \(f(t)\) ‍increases by \(\frac{\Delta t}{2.05}\).

  Therefore, for ‍\(f(t)\) to increase by ‍1, we need ‍\(\Delta t\). In other words, the ‍\(t\)-interval we are looking for is 2.05 ‍months.

• Barney is an ecologist who studies the change in the tiger population of Siberia over time.

  The relationship between the elapsed time ‍\(t\), in decades, since Barney started studying the population, and the number of tigers, ‍\(N(t)\), is modeled by the following function:

  \(N(t) = 710 \cdot (\frac{8}{125})^{t}\)

  The tiger population loses \(\frac{3}{5}\) of its size every 0.33 decades.

  Here is why.

  Note that losing \(\frac{3}{5}\) of the population is the same as being multiplied by \(\frac{2}{5}\).

  So we want to find the unit interval over which the tiger population is multiplied by \(\frac{2}{5}\).

  Let's rewrite the exponent so that the base equals \(\frac{2}{5}\). For this, we need to rewrite \(\frac{8}{125}\) as a power of \(\frac{2}{5}\).

  \(\left(\dfrac{8}{125}\right)^{ t}=\left(\left(\dfrac{2}{5}\right)^{3}\right)^t=\left(\dfrac{2}{5}\right)^{3t}\)

  Therefore, we can rewrite the modeling function as \(N(t)=710\cdot \left(\dfrac{2}5\right)^{ 3t}\).

  The modeling function is now written in the form \(A\cdot B^{ {f(t)}}\) where \(B=\dfrac{2}5\) and \(f(t)=3t\).

  The tiger population loses \(\dfrac{3}5\) of its size each time f(t) increases by 1, which happens each time \(t\) increases by \({\dfrac{1}{3}}\) or 0.33 (rounded to two decimal places).

• Gottfried wanted to see how contagious yawning can be. To better understand this, he conducted a social experiment by yawning in front of a random large crowd and observing how many people yawned as a result.

  The relationship between the elapsed time ‍\(t\), in minutes, since Gottfried yawned, and the number of people in the crowd, ‍\(P_{\text{minute}}(t)\), who yawned as a result is modeled by the following function:

  \(P_{\text{minute}}(t)=5\cdot (1.03)^{\Large t}\)

  Every hour, the number of people who yawn in Gottfried's experiment grows by a factor of 5.89.

  Here is why.

  We want to find the expression for \(P_{\text{hour}}(x)\), which models the number of people who yawn in Gottfried's experiment after ‍\(x\) hours.

  Since there are ‍60 minutes in an hour, ‍\(t = 60x\), and the expression for \(P_{\text{hour}}(x)\) ‍is the same as the following expression.

  \(P_{\text{minute}}(60x) =5\cdot \left(1.03\right)^{\Large 60x}=5\cdot \left((1.03)^{60}\right)^{\Large x}\)

  Evaluating ‍\((1.03)^{60}\) and rounding to two decimal places, we find that \(P_{\text{hour}}(x) =5\cdot \left(5.89\right)^{\Large x}\).

Construct Exponential Models

Here are examples.

• Tobias sent a chain letter to his friends, asking them to forward the letter to more friends. The number of people who receive the email increases by a factor of ‍4 every 9.1 ‍weeks, and can be modeled by a function, ‍\(P\), which depends on the amount of time, \(t\) ‍(in weeks).

  Tobias initially sent the chain letter to ‍37 friends.

  \(P(t) = 37 \cdot 4^{\frac{t}{9.1}}\)

  Here is why.

  We can model the situation with an exponential function of the general form \(A \cdot B^{f(t)}\), where \(A\) (37) ‍is the initial quantity, \(B\) (4) ‍is a factor by which the quantity is multiplied over constant time intervals, and \(f(t)\) ‍is an expression in terms of ‍that determines those time intervals.

  We know that the number of people who receive the email is multiplied by ‍4 every 9.1 ‍weeks. This means that each time ‍\(t\) increases by ‍9.1, \(f(t)\) ‍increases by ‍1. Therefore, \(f(t)\) ‍is a linear function whose slope is \(\frac{1}{9.1}\).

• Carbon-‍14 is an element which loses exactly half of its mass every 5730 ‍years. The mass of a sample of carbon-‍can be modeled by a function, ‍\(M\), which depends on its age, \(t\) ‍(in years).

  We measure that the initial mass of a sample of carbon-‍14 is ‍741 grams.

  \(M(t) = 741 \cdot 0.5^{\frac{t}{5730}}\)

  Here is why.

  We can model the situation with an exponential function of the general form \(A \cdot B^{f(t)}\), where \(A\) (741) ‍is the initial quantity, \(B\) (0.5 / half) ‍is a factor by which the quantity is multiplied over constant time intervals, and \(f(t)\) ‍is an expression in terms of ‍that determines those time intervals.

  We know that the mass of the sample is multiplied by ‍0.5 every 5730 ‍years. This means that each time ‍\(t\) increases by ‍5730, \(f(t)\) ‍increases by ‍1. Therefore, \(f(t)\) ‍is a linear function whose slope is \(\frac{1}{5730}\).

• Bela started studying how the number of branches on her tree grows over time. Every ‍2.9 years, the number of branches increases by an additional ‍83%, and can be modeled by a function, ‍\(N\), which depends on the amount of time, ‍\(t\) (in years).

  When Bela began the study, her tree had ‍60 branches.

  \(N(t) = 60 \cdot 1.83^{\frac{t}{2.9}}\)

  Here is why.

  We can model the situation with an exponential function of the general form \(A \cdot B^{f(t)}\), where \(A\) (60) ‍is the initial quantity, \(B\) (1.83 / increasing 83%) ‍is a factor by which the quantity is multiplied over constant time intervals, and \(f(t)\) ‍is an expression in terms of ‍that determines those time intervals.

  We know that the number of branches is multiplied by ‍1.83 every 2.9 ‍years. This means that each time ‍\(t\) increases by ‍2.9, \(f(t)\) ‍increases by ‍1. Therefore, \(f(t)\) ‍is a linear function whose slope is \(\frac{1}{2.9}\).

Logarithms

Logarithms are the inverse of exponential functions, used to find the exponent to which a base must be raised to produce a given number. For example, since \(10^{2}=100\), the \(\log _{10}100 = \log{100} = 2\).

Here are examples.

• \(\log_{2}{16} = 4\) because \(2^4 = 2 \cdot 2 \cdot 2 \cdot 2 = 16\)
• \(\log_{3}{81} = 4\) because \(3^4 = 3 \cdot 3 \cdot 3 \cdot 3 = 81\)
• \(\log_{5}{\frac{1}{125}} = -3\) because \(5^{-3} = 5^{-1} \cdot 5^{-1} \cdot 5^{-1} = \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} = \frac{1}{125}\)

\(e\) and Natural Logarithm

\[e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n} = \sum_{k=0}^{\infty} \frac{1}{k!} \approx 2.718281828 \ldots\]

The constant \(e\), also known as Euler's number, is an irrational mathematical constant approximately equal to \(2.71828\). It is the base of the natural logarithm and is fundamental to calculus and the study of exponential growth and decay.

\[\log_{e} = \ln\]

Here is an example of evaluating \(\log_{e}\) or \(\ln\). To prove this example, you need a Scientific Calculator.

\(\log_{e}{67} = \ln{67} \approx 4.205\)

Logarithm Properties

• Product rule: \(\log_b(MN)=\log_b(M)+\log_b(N)\)
• Quotient rule: \(\log_b\left(\dfrac{M}{N}\right)=\log_b(M)-\log_b(N)\)
• Power rule: \(\log_b(M^p)=p\cdot\log_b(M)\)
• Change of base rule: \(log_{b}(x) = \Large{\frac{log_{a}(x)}{log_{a}(b)}}\)
• Log of 1: \(log_{b}(1)=0\)
• Log of Base: \(log_{b}(b)=1\)
• Inverse Properties: \(\Large{log_{b}(b^{x}) = b^{log_{b}(x)}}\)

Expansion of Logarithmic Expressions

Expansion is the process of breaking a single, complex logarithm into multiple, simpler logarithms using the properties. This results in a longer expression. A general strategy is to apply the rules in the order of the quotient rule, then the product rule, and finally the power rule.

Here is an example.

\(\log _{b}\left(\frac{6x}{y}\right) = \log _{b}(6x)-\log _{b}(y) = \log _{b}(6)+\log _{b}(x)-\log _{b}(y)\)

Compression of Logarithmic Expressions

Compression (or condensing) is the reverse process, where multiple logarithmic terms are combined into a single logarithm. This is done by reversing the properties, often using the power rule first, followed by the product and quotient rules.

Here is an example.

\(\ln (x^{2})+\ln (x+1) = \ln (x^{2}\times (x+1)) = \ln (x^{3}+x^{2})\)

Solving Exponential Equations and Models using Logarithms

We just learned about e, Logarithm properties, and Expansion/Compression of Logarithmic Expressions. Let's solve some exponential equations and models.

• \begin{aligned} 2 \cdot 6^{x} &= 236 \\\\ 6^{x} &= \frac{236}{2} \\\\ 6^{x} &= 118 \\\\ x &= \log_{6}{118} \\\\ x &= \frac{\log{118}}{\log{6}} \\\\ x &\approx 2.662 \end{aligned}
• \begin{aligned} 6 \cdot e^{y} &= 300 \\\\ e^{y} &= \frac{300}{6} \\\\ e^{y} &= 50 \\\\ y &= \log_{e}{50} \\\\ y &= \ln{50} \\\\ y &\approx 3.912 \end{aligned}

• Noah borrows ‍$2000 from his father and agrees to repay the loan and any interest determined by his father as soon as he has the money.

  The relationship between the amount of money, ‍\(A\), in dollars that Noah owes his father (including interest), and the elapsed time, ‍\(t\), in years, is modeled by the following equation.

  \(A=2000e^{0.1t}\)

  How long did it take Noah to pay off his loan if the amount he paid to his father was equal to ‍$2450?

  Here is the answer.

  \begin{aligned} A &= 2000e^{0.1t} \\\\ 2450 &= 2000e^{0.1t} \\\\ \frac{2450}{2000} &= e^{0.1t} \\\\ \frac{49}{40} &= e^{0.1t} \\\\ t &= \frac{\ln{\frac{49}{40}}}{0.1} \\\\ t &= \frac{\ln{1.225}}{0.1} \\\\ t &\approx 2 \end{aligned}

Reference

The primary reference is Khan Academy, with assistance from ChatGPT and Gemini as well.

Related Articles

• Algebra (Part 4)
• Algebra (Part 3)
• Algebra (Part 2)
• Algebra (Part 1)
• Pre-Algebra

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