Logarithms with Exponents - Algebra II

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Question

Simplify the following equation.

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

Answer

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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Question

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 .

Answer

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:

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

or

$\small x \approx 2.2620$

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Question

Simplify

$\log6^2$

Answer

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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Question

Evaluate the following expression

$\left ( \log1000\right)^{2}=$

Answer

This is a simple exponent of a logarithmic answer.

$\small \log(1000)=3$ because $\small 10^{3}=1000$

$\small 3^{2}=9$

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Question

Evaluate the following expression

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

Answer

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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Question

Evaluate the following expression

$\left( \log _{2}16 \right)^{3}$

Answer

This is a two step problem. First find the log base 2 of 16

$\small \log _{2}(16)=4$ because $\small 2^{4}=16$

then

$\small 4^{3}=64$

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Question

Evaluate the following expression

$\small \log _{3}(9^{2})$

Answer

Since the exponent is inside the parentheses, you must determine the value of the exponential expression first.

$\small 9^{2}=81$

then you solve the logarithm

$\small \log _{3}(81)=4$ because $\small 3^{4}=81$

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Question

Which of the following equations is valid?

Answer

Since a logarithm answers the question of which exponent to raise the base to receive the number in parentheses, if the number in parentheses is the base raised to an exponent, the exponent must be the answer.

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Question

Evaluate the following for all integers of $\small a$ and $\small b$

$\small a^{\log _{a}(b)}$

Answer

$\small \log _{a}(b)$ gives us the exponent to which $\small a$ must be raised to yield $\small b$

When $\small a$ is actually raised to that number in the equation given, the answer must be $\small b$

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Question

Rewrite the following logarithmic expression into expanded form (that is, using addition and/or subtraction):

$\small \ln{(AB)}^{4}$

Answer

Before we do anything, the exponent of 4 must be moved to the front of the expression, as the rules of logarithms dictate. We end up with $\small 4\ln{AB}$ . Remember that a product inside of a logarithm can be rewritten as a sum: $\small 4(\ln{A}+\ln{B})$ . Distributing, we get $\small 4\ln{A}+4\ln{B}$ .

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Question

Use

$\log4=0.602$

and

$\log12=1.079$

Evaluate:

$\log64$

Answer

Since the question gives,

$\log4=0.602$

and

$\log12=1.079$

To evaluate

$\log64$

manipulate the expression to use what is given.

$=\log4^3$

$=3\log4$

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Question

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

Answer

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:

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Question

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

Answer

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:

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:

The answer is:

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Question

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

Answer

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:

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Question

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

Answer

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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Question

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

Answer

By the Logarithm of a Power Property, for all real , all ,

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

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 for which the expressions are defined,

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

setting , th equation becomes

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

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Question

Which statement is true of

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

for all integers ?

Answer

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 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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