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Algebra II : Logarithms

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Example Questions

Example Question #42 : Adding And Subtracting Logarithms

Simplify 2 log (x) - log ( x + 1) .

Possible Answers:

$log(\frac{x^2}{x+1})$

log(x^2-x-1)

log(x^2-x+1)

log(x^2+x-1)

log(x^2+x+1)

Correct answer:

$log(\frac{x^2}{x+1})$

Explanation:

First we can make the coefficient from the left term into an exponent:

log (x^2) - log ( x + 1)

Next, remember that if we're subtracting logs, we divide the terms inside them:

$log(\frac{x^2}{x+1})$

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Example Question #41 : Adding And Subtracting Logarithms

Simplify log (2x) + log (x - 3)

Possible Answers:

log(3x-3)

log(2x^2-6x)

log(x^2+x-3)

2log(x^2-3x)

$log(\frac{2x}{x-3})$

Correct answer:

log(2x^2-6x)

Explanation:

When we add logs, we multiply the terms in them:

log((2x)(x-3))

From here, we multiply them out:

log(2x^2-6x)

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Example Question #54 : Simplifying Logarithms

True or false: for all positive values of ,

$\log 100N = 2+\log N$

Possible Answers:

True

False

Correct answer:

True

Explanation:

By the Product of Logarithms Property,

$\log MN = \log M +\log N$

Setting M = 100 , this becomes

$\log 100N = \log 100 +\log N$

"log" refers to the common, or base ten, logarithm, so, by definition,

$\log M = a$

if and only if

$10^{a} = M$ .

Setting M = 100

$10^{a} = 100$

$10^{a} = 10^{2}$

a =2 ,

so $\log 100 = 2$

and $\log 100N = \log 100 +\log N = 2 + \log N$ .

The statement is true.

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Example Question #55 : Simplifying Logarithms

True or false: $\log e ^{N} = N$ for all values of .

Possible Answers:

False

True

Correct answer:

False

Explanation:

A statement can be proved to not be true in general if one counterexample can be found. One such counterexample assumes that N = 1 . The statement

$\log e ^{N} = N$

becomes

$\log e ^{1} = 1$

or, equivalently,

$\log e = 1$

The word "log" indicates a common, or base ten, logarithm, as opposed to a natural, or base , logarithm. By definition, the above statement is true if and only if

$10 ^{1} = e$ ,

10= e .

This is false, so $\log e ^{N} = N$ does not hold for N = 1 . Since the statement fails for one value, it fails in general.

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Example Question #42 : Adding And Subtracting Logarithms

True or false:

$\log\left ( -2 N \right ) = \log (-2) + \log N$

for all negative values of .

Possible Answers:

False

True

Correct answer:

False

Explanation:

It is true that by the Product of Logarithms Property,

$\log MN = \log M +\log N$ .

However, this only holds true if both and are positive. The logarithm of a negative number is undefined, so the expression $\log (-2)$ is undefined. The statement is therefore false.

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Example Question #43 : Adding And Subtracting Logarithms

True or false: $\log _{7} N = \frac{\log_{6}N}{\log_{6}7}$ for all positive .

Possible Answers:

False

True

Correct answer:

True

Explanation:

By the Change of Base Property of Logarithms, if M,N,a > 0 and $a, M \ne 1$ ,

$\log_{M}N = \frac{\log_{a}N}{\log_{a}M}$

Substituting 7 for and 6 for , the statement becomes the given statement

$\log _{7} N = \frac{\log_{6}N}{\log_{6}7}$ .

The correct choice is "true."

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Example Question #121 : Logarithms

In this question we will use the notation $\small \log a$ to represent the base 10 or common logarithm, i.e. $\small \log a \equiv \log_{10}a$ .

Find if 3^x=12 .

Possible Answers:

$\small x \approx 0.4771$

$\small x \approx 2.9891$

$\small x \approx 4.000$

$\small x \approx 2.2620$

$\small x \approx 2.3116$

Correct answer:

$\small x \approx 2.2620$

Explanation:

We can use the Property of Equality for Logarithmic Functions to take the logarithm of both sides:

$\small \log 3^x = \log 12$

Use the Power Property of Logarithms:

$\small x \log 3 = \log 12$

Divide each side by $\log3$ :

$\small x=\frac{\log 12}{\log 3}$

Use a calculator to get:

$\small x \approx \frac{1.0792}{0.4771}$

$\small x \approx 2.2620$

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Example Question #122 : Logarithms

Simplify

$\log6^2$

Possible Answers:

$\log12$

$2\log6$

$(\log6)^2$

$\log3$

Correct answer:

$2\log6$

Explanation:

Using Rules of Logarithm recall:

$\log a^b = b\log a$ .

Thus, in this situation we bring the 2 in front and we get our solution.

$2 \log(6)$

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Example Question #125 : Logarithms

Simplify the following equation.

$(e^{(\ln2+\ln93-\ln6)}e^{-1})^2$

Possible Answers:

31e

$\frac{31}{e^{-2}}$

961e^2

$\frac{961}{e^2}$

$\frac{31}{e}$

Correct answer:

$\frac{961}{e^2}$

Explanation:

We can simplify the natural log exponents by using the following rules for naturla log.

$\ln A+\ln B=\ln AB$

$\ln A - \ln B=\ln\frac{A}{B}$

Using these rules, we can perform the following steps.

$(e^{(\ln2+\ln93-\ln6)}e^{-1})^2$

$(e^{(\ln(2(93))-\ln6)}e^{-1})^2$

$(e^{(\ln(\frac{2(93)}{6})}e^{-1})^2$

$(e^{\ln 31}e^{-1})^2$

Knowing that the e cancels the exponential natural log, we can cancel the first e.

$(31e^{-1})^2$

Distribute the square into the parentheses and calculate.

$(31)^2(e^{-1})^2=961e^{-2}$

Remember that a negative exponent is equivalent to a quotient. Write it as a quotient and then you're finished.

$961e^{-2}=\frac{961}{e^2}$

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Example Question #1 : Logarithms With Exponents

Evaluate the following expression

$\small \log(1000^{2})$

Possible Answers:

$\small \frac{\log (2)}{\log (10)}$

$\small 9$

$\small 3$

$\small 6$

$\small \frac{1}{9}$

Correct answer:

$\small 6$

Explanation:

Since the exponent is inside the parentheses, you must take the square of 1000 before finding the logarithim. Therefore

$\small \log(1000^{2})=\log (1000000)=6$

because $\small 10^{6}=1,000,000$

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Algebra II : Logarithms

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #42 : Adding And Subtracting Logarithms

Example Question #41 : Adding And Subtracting Logarithms

Example Question #54 : Simplifying Logarithms

Example Question #55 : Simplifying Logarithms

Example Question #42 : Adding And Subtracting Logarithms

Example Question #43 : Adding And Subtracting Logarithms

Example Question #121 : Logarithms

Example Question #122 : Logarithms

Example Question #125 : Logarithms

Example Question #1 : Logarithms With Exponents

All Algebra II Resources

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