Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #13 : Solve Logarithmic Equations

Solve the following logarithmic equation.  

\(\displaystyle log_{2}\,x +log_{2}\,(4-x)=2\) 

 

Possible Answers:

\(\displaystyle x=1\)

\(\displaystyle x=2\)

\(\displaystyle x=-4\)

\(\displaystyle x=0\)

Correct answer:

\(\displaystyle x=2\)

Explanation:

In order to solve the logarithmic equation, we use the following property

  • \(\displaystyle log\,a+log\,b=log\,ab\)

As such

\(\displaystyle log_{2}\,x +log_{2}\,(4-x)=log_{2}\,x(4-x)=log_2\,4x-x^2=2\)

And converting from logarithmic form to exponential form

\(\displaystyle log_a\,c=b \Rightarrow a^b=c\)

we get

\(\displaystyle 2^2=4x-x^2\)

Solving for x

\(\displaystyle x^2-4x+4=0\)

And because the square of a difference is given as this equation through factoring

\(\displaystyle (a-b)^2=a^2-2ab+b^2\)

we have

\(\displaystyle (x-2)^2=0\)

which implies

\(\displaystyle x=2\)

Example Question #11 : Solve Logarithmic Equations

Express the log in its expanded form:

\(\displaystyle \log\bigg(\frac{x^3y}{z}\bigg)\)

Possible Answers:

\(\displaystyle \log(x)+\log(y)-\log(z)\)

\(\displaystyle 3\log(x)+\log(y)- \log(z)\)

\(\displaystyle 3\log(x)+\log(y)+\log(z)\)

None of the other answers

\(\displaystyle \log^{3}(x)+\log(y)-\log(z)\)

Correct answer:

\(\displaystyle 3\log(x)+\log(y)- \log(z)\)

Explanation:

You need to know the Laws of Logarithms in order to solve this problem. The ones specifically used in this problem are the following:

\(\displaystyle \log(xy) = \log (x)+\log(y)\)

\(\displaystyle \log\bigg(\frac{x}{y}\bigg) = \log(x)-\log(y)\)

\(\displaystyle \log(x^y) = y\cdot \log(x)\)

Let's take this one variable at a time starting with expanding z:

\(\displaystyle \log\bigg(\frac{x^3y}{z}\bigg) = \log(x^3y)-\log(z)\)

Now y:

\(\displaystyle \log(x^3y)-\log(z) = \log(x^3)+\log(y)-\log(z)\)

And finally expand x:

\(\displaystyle \log(x^3)+\log(y)-\log(z) = 3\log(x)+\log(y)-\log(z)\)

Example Question #11 : Solve Logarithmic Equations

What is \(\displaystyle ln(2x)-ln(2)\) equivalent to?

Possible Answers:

\(\displaystyle \ln (2x-2)\)

\(\displaystyle \ln(2)\)

\(\displaystyle \ln(x)\)

\(\displaystyle \frac{\ln(2x)}{2}\)

Correct answer:

\(\displaystyle \ln(x)\)

Explanation:

Using the properties of logarithms,

\(\displaystyle \log(a)-\log(b)=\log \bigg( \frac{a}{b}\bigg)\)

the expression can be rewritten as

 \(\displaystyle \ln(2x)-\ln(2)=\ln\bigg(\frac{2x}{2}\bigg)\)

 which simplifies to \(\displaystyle \ln(x)\).

Example Question #12 : Solve Logarithmic Equations

Find the value of the sum of logarithms by condensing the expression. 

\(\displaystyle \ln(2)+\ln\bigg(\frac{1}{2}\bigg)\)

Possible Answers:

Undefined

\(\displaystyle e\)

\(\displaystyle 1\)

\(\displaystyle 0\)

Correct answer:

\(\displaystyle 0\)

Explanation:

By the property of the sum of logarithms,

\(\displaystyle \ln(2)+\ln\bigg(\frac{1}{2}\bigg)=\ln\bigg(2\cdot\frac{1}{2}\bigg)=\ln(1)=0\).

Example Question #81 : Exponential And Logarithmic Functions

Condense the following logarithmic equation:

\(\displaystyle 3\log (y)+\log (4)+\log (x)-2\log (z)\)

Possible Answers:

\(\displaystyle \log \left(\frac{4xy^3}{z^2}\right)\)

\(\displaystyle \log \left(\frac{xy^3}{2z}\right)\)

\(\displaystyle \log \left(\frac{z^2}{4xy^3}\right)\)

\(\displaystyle \log (4xy^3-z^2)\)

\(\displaystyle \log \left(\frac{2z^2}{xy^3}\right)\)

Correct answer:

\(\displaystyle \log \left(\frac{4xy^3}{z^2}\right)\)

Explanation:

We start condensing our expression using the following property, which allows us to express the coefficients of two of our terms as exponents:

\(\displaystyle b\log (a)=\log (a^b)\)

\(\displaystyle 3\log (y)+\log (4)+\log (x)-2\log (z)\)

\(\displaystyle \log(y^3)+\log(4)+\log(x)-\log(z^2)\)

Our next step is to use the following property to combine our first three terms:

\(\displaystyle \log (a)+\log(b)+\log(c)=\log(abc)\)

\(\displaystyle \log(y^3)+\log(4)+\log(x)-\log(z^2)\)

\(\displaystyle \log(4xy^3)-\log(z^2)\)

Finally, we can use the following property regarding subtraction of logarithms to obtain the condensed expression:

\(\displaystyle \log (a)-\log (b)=\log (\frac{a}{b})\)

\(\displaystyle \log(4xy^3)-\log(z^2)\)

\(\displaystyle \log \left(\frac{4xy^3}{z^2}\right)\)

Example Question #41 : Properties Of Logarithms

What is another way of writing

 \(\displaystyle \frac{\ln(x^b)}{\ln(y^a)}\)?

Possible Answers:

\(\displaystyle \frac{\ln(x^{1/a})}{\ln(y^{1/b})}\)

\(\displaystyle \frac{\ln(x^{b-a})}{\ln y}\)

\(\displaystyle \ln [(x/y)^{\frac{b}{a}}]\)

\(\displaystyle \ln [(x/y)^{b-a}]\)

Correct answer:

\(\displaystyle \frac{\ln(x^{1/a})}{\ln(y^{1/b})}\)

Explanation:

The correct answer is 

\(\displaystyle \frac{\ln(x^{1/a})}{\ln(y^{1/b})}\)

Properties of logarithms allow us to rewrite \(\displaystyle \ln y^a\) and \(\displaystyle \ln x^b\) as \(\displaystyle \ln y^a=a\ln y\) and \(\displaystyle b\ln x\), respectively. So we have

\(\displaystyle \frac{\ln(x^b)}{\ln(y^a)}=\frac{b}{a}\frac{\ln(x)}{\ln(y)}=\frac{1/a}{1/b}\frac{\ln(x)}{\ln(y)}\)

Again, we use the logarithm property

 \(\displaystyle \frac{1}{a}\ln x=\ln x^{1/a}\) 

to get

\(\displaystyle \frac{\ln(x^b)}{\ln(y^a)}=\frac{1/a}{1/b}\frac{\ln(x)}{\ln(y)}=\frac{\ln(x^{1/a})}{\ln(y^{1/b})}\)

Example Question #19 : Solve Logarithmic Equations

Write the expression in the most condensed form:\(\displaystyle 4\ln(x)-\frac{1}{2}\ln(x+4)\)

Possible Answers:

\(\displaystyle ln\frac{4x}{\sqrt{x+4}}\)

\(\displaystyle ln\frac{4x}{x+4}\)

\(\displaystyle ln\frac{x}{(x+4)^{\frac{1}{2}}}\)

None of the other answers.

\(\displaystyle ln\frac{x^4}{\sqrt{x+4}}\)

Correct answer:

\(\displaystyle ln\frac{x^4}{\sqrt{x+4}}\)

Explanation:

\(\displaystyle 4\ln(x)-\frac{1}{2}\ln(x+4)\)

Use the Power property of Logarithms:

\(\displaystyle \ln(x)^4-\ln(x+4)^{\frac{1}{2}}\)

Rewrite the fractional exponent:

\(\displaystyle \ln(x)^4-\ln\sqrt{(x+4)}\)

Condense into a fraction using the Quotient property of Logarithms:

\(\displaystyle \mathbf{\ln\frac{x^4}{\sqrt{x+4}}}\)

Example Question #20 : Solve Logarithmic Equations

Simplify:  \(\displaystyle \log_2 (8)-\log_2(15)+\log_2(10)\)

Possible Answers:

\(\displaystyle \log(\frac{16}{3})\)

\(\displaystyle \log_2(\frac{16}{3})\)

\(\displaystyle \log_2(3)\)

\(\displaystyle \log(3)\)

\(\displaystyle \log_6(3)\)

Correct answer:

\(\displaystyle \log_2(\frac{16}{3})\)

Explanation:

When logs of the same bases are subtracted, the contents of both logs will be divided with each other.  When logs of the same bases are added, then the contents inside the log will be multiplied together.

\(\displaystyle \log_2 (8)-\log_2(15)= \log_2(8\div 15)=\log_2(\frac{8}{15})\)

\(\displaystyle \log_2(\frac{8}{15}) + \log_2(10)=\log_2(\frac{8}{15}\times 10) = \log_2(\frac{16}{3})\)

Example Question #21 : Solve Logarithmic Equations

Completely condense the logarithm: \(\displaystyle \frac{1}{3}log_{10}x-\frac{1}{2}[log_{10}(x+y)]\).

Possible Answers:

\(\displaystyle log_{10}\frac{\sqrt{x}}{\sqrt{xy}}\)

None of the other answers.

\(\displaystyle log_{10}\frac{\sqrt[3]{x}}{\sqrt{x+y}}\)

\(\displaystyle log_{10}\frac{\sqrt{x}}{\sqrt{x+y}}\)

\(\displaystyle log_{10}\frac{\sqrt[3]{x}}{\sqrt{xy}}\)

Correct answer:

\(\displaystyle log_{10}\frac{\sqrt[3]{x}}{\sqrt{x+y}}\)

Explanation:

\(\displaystyle \frac{1}{3}log_{10}x-\frac{1}{2}[log_{10}(x+y)]\)

Apply the Power property:

\(\displaystyle log_{10}x^\frac{1}{3}-log_{10}(x+y)^\frac{1}{2}\)

\(\displaystyle log_{10}\sqrt[3]{x}-log_{10}\sqrt{x+y}\)

Apply the quotient property:

\(\displaystyle \mathbf{log_{10}\frac{\sqrt[3]{x}}{\sqrt{x+y}}}\)

Example Question #21 : Solve Logarithmic Equations

Simplify

\(\displaystyle \small \log \,3\,\,+\,\,\log\,7\)

Possible Answers:

\(\displaystyle \small \log\,\frac{3}{7}\)

\(\displaystyle \small \log\,21\)

\(\displaystyle \small \log\,10\)

\(\displaystyle \small \log\,37\)

Correct answer:

\(\displaystyle \small \log\,21\)

Explanation:

By the property of the addition of logarithms with the same base

\(\displaystyle \small \small \log\,a\,\,+\,\,\log\,b=\log\,ab\)

As such

\(\displaystyle \small \small \log \,3\,\,+\,\,\log\,7=\log\,3\cdot7=\log\,21\)

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