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Algebra II : Logarithms with Exponents

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Example Questions

Example Question #11 : Logarithms With Exponents

Use

$\log4=0.602$

and

$\log12=1.079$

Evaluate:

$\log64$

Possible Answers:

1.806

0.218

1.204

0.477

1.681

Correct answer:

1.806

Explanation:

Since the question gives,

$\log4=0.602$

and

$\log12=1.079$

To evaluate

$\log64$

manipulate the expression to use what is given.

$=\log4^3$

$=3\log4$

=3*0.602

=1.806

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Example Question #380 : Mathematical Relationships And Basic Graphs

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

Possible Answers:

4x+2

4x+1

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

$2^{2x+1}$

4x^2+4x+1

Correct answer:

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.

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

Distribute the integer to both terms of the binomial.

The answer is: 4x+2

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Example Question #12 : Logarithms With Exponents

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

Possible Answers:

25-x

5-x

125-5x

125-25x

x-5

Correct answer:

125-25x

Explanation:

The natural log has a default base of .

This means that the expression written can also be:

$5ln_e(e^{5(5-x)})$

Recall the log property that:

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 will be eliminated.

The expression will simplify to:

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

The answer is: 125-25x

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Example Question #131 : Logarithms

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

Possible Answers:

5x-6

25x^2+60x-36

10x-12

5x-4

25x^2-60x-36

Correct answer:

5x-6

Explanation:

The log property need to solve this problem is:

$b^{log_b x} = x$

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

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

The answer is: 5x-6

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Example Question #14 : Logarithms With Exponents

Solve: $log_{4}(8^3)$

Possible Answers:

$\frac{9}{2}$

$\frac{8}{3}$

Correct answer:

$\frac{9}{2}$

Explanation:

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

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

Both terms can now be rewritten in base two.

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

The exponents can be moved to the front as coefficients.

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

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

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Example Question #73 : Simplifying Logarithms

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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

setting b = 5 , th equation becomes

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

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Example Question #15 : Logarithms With Exponents

Which statement is true of

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

for all integers N > 1 ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Due to the following relationship:

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

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

can be rewritten as

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

By definition,

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

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

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

or, substituting back,

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

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Algebra II : Logarithms with Exponents

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #11 : Logarithms With Exponents

Example Question #380 : Mathematical Relationships And Basic Graphs

Example Question #12 : Logarithms With Exponents

Example Question #131 : Logarithms

Example Question #14 : Logarithms With Exponents

Example Question #73 : Simplifying Logarithms

Example Question #15 : Logarithms With Exponents

All Algebra II Resources

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