Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #11 : Logarithms With Exponents

Use 

\displaystyle \log4=0.602 

and 

\displaystyle \log12=1.079

Evaluate: 

\displaystyle \log64

Possible Answers:

\displaystyle 1.681

\displaystyle 0.477

\displaystyle 0.218

\displaystyle 1.806

\displaystyle 1.204

Correct answer:

\displaystyle 1.806

Explanation:

Since the question gives,

\displaystyle \log4=0.602 

and 

\displaystyle \log12=1.079

To evaluate 

\displaystyle \log64

manipulate the expression to use what is given.

\displaystyle =\log4^3

\displaystyle =3\log4

\displaystyle =3*0.602

\displaystyle =1.806

Example Question #380 : Mathematical Relationships And Basic Graphs

Simplify:  \displaystyle 2e^{ln(2x+1)}

Possible Answers:

\displaystyle 4x+2

\displaystyle 4x+1

\displaystyle x+\frac{1}{2}

\displaystyle 2^{2x+1}

\displaystyle 4x^2+4x+1

Correct answer:

\displaystyle 4x+2

Explanation:

According to log rules, when an exponential is raised to the power of a logarithm, the exponential and log will cancel out, leaving only the power.

Simplify the given expression.

\displaystyle 2e^{ln(2x+1)} = 2(2x+1)

Distribute the integer to both terms of the binomial.

The answer is:  \displaystyle 4x+2

Example Question #11 : Logarithms With Exponents

Simplify:  \displaystyle 5ln(e^{5(5-x)})

Possible Answers:

\displaystyle 125-25x

\displaystyle 25-x

\displaystyle 125-5x

\displaystyle x-5

\displaystyle 5-x

Correct answer:

\displaystyle 125-25x

Explanation:

The natural log has a default base of \displaystyle e.

This means that the expression written can also be:

\displaystyle 5ln_e(e^{5(5-x)})

Recall the log property that: 

\displaystyle ln_k(k^n) = n

This would eliminate both the natural log and the base, leaving only the exponent.

The natural log and the base \displaystyle e will be eliminated.

The expression will simplify to:

\displaystyle 5[5(5-x)] = 5[25-5x] = 125-25x

The answer is:  \displaystyle 125-25x

Example Question #11 : Logarithms With Exponents

Simplify:  \displaystyle 2^{log_2(5x-6)}

Possible Answers:

\displaystyle 5x-4

\displaystyle 10x-12

\displaystyle 25x^2+60x-36

\displaystyle 25x^2-60x-36

\displaystyle 5x-6

Correct answer:

\displaystyle 5x-6

Explanation:

The log property need to solve this problem is:

\displaystyle b^{log_b x} = x

The base and the log of the base are similar.  They will both cancel and leave just the quantity of log based two.

\displaystyle 2^{log_2(5x-6)}=5x-6

The answer is:  \displaystyle 5x-6

Example Question #14 : Logarithms With Exponents

Solve:  \displaystyle log_{4}(8^3)

Possible Answers:

\displaystyle 9

\displaystyle \frac{9}{2}

\displaystyle \frac{8}{3}

\displaystyle 8

\displaystyle 4

Correct answer:

\displaystyle \frac{9}{2}

Explanation:

Rewrite the log so that the terms are in a fraction.

\displaystyle log_{4}(8^3) = \frac{log(8^3)}{log( 4)}

Both terms can now be rewritten in base two.

\displaystyle \frac{log(8^3)}{log( 4)} = \frac{log((2^3)^3)}{log( 2^2)} = \frac{log(2)^9}{log(2)^2}

The exponents can be moved to the front as coefficients.

\displaystyle \frac{log(2)^9}{log(2)^2} = \frac{9log(2)}{2log(2)} = \frac{9}{2}

The answer is:  \displaystyle \frac{9}{2}

Example Question #381 : Mathematical Relationships And Basic Graphs

Which statement is true of \displaystyle \frac{1}{5} \log _{5}N for all positive values of \displaystyle N?

Possible Answers:

\displaystyle \frac{1}{5} \log _{5}N = \log_{5} (N - 1)

\displaystyle \frac{1}{5} \log _{5}N = \log_{5}\left ( \frac{1}{5} N \right )

\displaystyle \frac{1}{5} \log _{5}N = \log_{5} \sqrt[5]{N}

\displaystyle \frac{1}{5} \log _{5}N = \log_{5} (N - 5)

\displaystyle \frac{1}{5} \log _{5}N = N

Correct answer:

\displaystyle \frac{1}{5} \log _{5}N = \log_{5} \sqrt[5]{N}

Explanation:

By the Logarithm of a Power Property, for all real \displaystyle b, all \displaystyle N > 0\displaystyle a > 0 , a \ne 1

\displaystyle b \log_{a} N = \log_{a}( N ^{b})

Setting \displaystyle b = \frac{1}{5} , a = 5, the above becomes 

\displaystyle \frac{1}{5} \log _{5}N = \log_{5} \left ( N^{\frac{1}{5}}\right )

Since, for any \displaystyle a,b for which the expressions are defined, 

\displaystyle N^{\frac{1}{b}} = \sqrt[b]{N},

setting \displaystyle b = 5, th equation becomes

\displaystyle \frac{1}{5} \log _{5}N = \log_{5} \sqrt[5]{N}.

Example Question #71 : Simplifying Logarithms

Which statement is true of 

\displaystyle \log_{7} \sqrt[N]{7}

for all integers \displaystyle N > 1?

Possible Answers:

\displaystyle \log_{7} \sqrt[N]{7}= \frac{1}{N}

\displaystyle \log_{7} \sqrt[N]{7}= \frac{7}{N}

\displaystyle \log_{7} \sqrt[N]{7} = -N

\displaystyle \log_{7} \sqrt[N]{7} = N - 1

\displaystyle \log_{7} \sqrt[N]{7} = N - 7

Correct answer:

\displaystyle \log_{7} \sqrt[N]{7}= \frac{1}{N}

Explanation:

Due to the following relationship:

\displaystyle \sqrt[N]{7} = 7 ^{\frac{1}{N}}; therefore, the expression 

\displaystyle \log_{7} \sqrt[N]{7}

can be rewritten as 

\displaystyle \log_{7} \left ( 7 ^{\frac{1}{N}} \right )

By definition,  

\displaystyle \log_{a} \left (a^{M} \right )= M.

Set \displaystyle a = 7 and \displaystyle M = \frac{1}{N}, and the equation above can be rewritten as

\displaystyle \log_{7} \left ( 7 ^{\frac{1}{N}} \right )= \frac{1}{N},

or, substituting back,

\displaystyle \log_{7} \sqrt[N]{7}= \frac{1}{N}

Example Question #1 : Solving Logarithms

Solve for \displaystyle x:

\displaystyle 5^{x} = \left(\frac{1}{25}\right)^{4x-1}

Possible Answers:

\displaystyle x = \frac{2}{9}

\displaystyle x = \frac{-2}{7}

\displaystyle x = \frac{-1}{2}

\displaystyle x = \frac{1}{3}

\displaystyle x = \frac{1}{2}

Correct answer:

\displaystyle x = \frac{2}{9}

Explanation:

To solve for \displaystyle x, first convert both sides to the same base:

\displaystyle 5^{x} = \frac{1}{25}^{4x-1} = (5^{-2})^{4x-1} = 5^{-8x+2}

Now, with the same base, the exponents can be set equal to each other:

\displaystyle x= - 8x +2

Solving for \displaystyle x gives:

\displaystyle x = \frac{2}{9}

Example Question #3041 : Algebra Ii

Solve the equation: 

\displaystyle \log{_{4}}{x^{2}}+\log{_{4}}{3}=1

Possible Answers:

\displaystyle 1,-1

\displaystyle \frac{2\sqrt{3}}{3}, -\frac{2\sqrt{3}}{3}

\displaystyle \frac{\sqrt{6}}{3}, -\frac{\sqrt{6}}{3}

\displaystyle \frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}

\displaystyle \frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{3}

Correct answer:

\displaystyle \frac{2\sqrt{3}}{3}, -\frac{2\sqrt{3}}{3}

Explanation:

\displaystyle \log{_{4}}{x^{2}}+\log{_{4}}{3}=1

\displaystyle \Rightarrow \log{_{4}}{x^{2}}+\log{_{4}}{3}=\log{_{4}}{4}

\displaystyle \Rightarrow log{_{4}}{(x^{2}\times 3)}=\log{_{4}}{4}

\displaystyle \Rightarrow 3x^{2}=4

\displaystyle \Rightarrow x^{2}=\frac{4}{3}

\displaystyle \Rightarrow x=\frac{2\sqrt{3}}{3}, x=-\frac{2\sqrt{3}}{3}

Example Question #3 : Solving Logarithms

Use  \displaystyle \small \small \log_23\approx 1.5850  to approximate the value of  \displaystyle \small \log_2 24.

Possible Answers:

\displaystyle \small \log_2 24\approx 3.4150

\displaystyle \small \log_2 24\approx 12.5850

\displaystyle \small \log_2 24\approx 3.5850

\displaystyle \small \log_2 24\approx 4.5850

\displaystyle \small \log_2 24\approx 12.6800

Correct answer:

\displaystyle \small \log_2 24\approx 4.5850

Explanation:

Rewrite \displaystyle 24 as a product that includes the number \displaystyle 3:

\displaystyle \small \log_2 24 = \log_2 (2^3 \cdot 3)

Then we can split up the logarithm using the Product Property of Logarithms:

\displaystyle \small \log_2 (2^3\cdot 3) = \log_2 2^3 + \log_2 3

                     \displaystyle \small =3+\log_2 3

                     \displaystyle \small \approx 3 + 1.5850

Thus,

\displaystyle \small \log_2 24 \approx 4.5850.

 

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