Logarithms - Algebra II

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Question

Simplify the expression using logarithmic identities.

Answer

The logarithm of a fraction is equal to the logarithm of the numerator minus the logarithm of the denominator.

$log_A(\frac{B}{C})=log_AB-log_AC$

If we encounter two logarithms with the same base, we can likely combine them. In this case, we can use the reverse of the above identity.

$log_AB-log_AC=log_A(\frac{B}{C})$

$A=4,\ B=7,\ C=5$

$log_47-log_45=log_4(\frac{7}{5})$

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Question

Simplify the following logarithmic expression:

$log_{3}81 - log_{5} (\frac{1}{25}) + log_{7} 7^{\frac{3}{2}}$

Answer

Each term can be simplified as follows:

$log_{3}81 = log_{3} 3^{4} = 4$

$- log_{5} (\frac{1}{25}) = - log_{5} (5)^{-2} = -(-2) = 2$

$log_{7} 7^{\frac{3}{2}} = \frac{3}{2}$

Combining these gives the answer: $\frac{15}{2}$

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Question

Use logarithmic properties to simplify this expression:

$log(4x)+log\left(\frac{1}{2}\right)-log(y^{2})$

Answer

Use the sum/product rule to combine the first 2 terms:

$log\left(4x\ast \frac{1}{2}\right)-log(y^{2})=log(2x)-log(y^{2})$

Use the difference/quotient rule to combine the remaining terms:

$log\left(\frac{2x}{y^{2}}\right)$

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Question

What is the value of ?

Answer

Remember the rules of logarithms:

$log_pq - log_pr = log_p\frac{q}{r}$

This means we can simplify it as follows:

$log_7(343\cdot 49) - log_72401$

$log_7\frac{16807}{2401}$

The logarithm of anything with the same base is always , so the correct answer is .

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Question

Expand the following logarithmic expression into a list of sums or subtractions of logarithms:

${}\log(\frac{a^3b^2}{c^4})$

Answer

One important property of logarithms is that multiplication inside the logarithm is the same thing as addition outside of it. In the same way division is "the same" as subtraction in logarithms. So our expression is the same as

${} \log a^3 + \log b^2 - \log c^4$

But also, exponents can be moved outside in the same way. is basically $a \cdot a \cdot a$ , so . This can be reduced even further to our final answer:

$3 \log a+2 \log b-4 \log c$

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Question

Which is another way of expressing

$\small \log(4)+\log(5)$ ?

Answer

Use the rule:

$\small \log(a)+\log(b)=\log(a\times b)$

therefore

$\small \log (4)+\log (5)=\log (4\times 5)=\log (20)$

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Question

Which of the following is another way to express

$\small \log (8)-\log(4)$ ?

Answer

Use the rule

$\small \log(a)-\log (b)=\log(\frac{a}{b})$

therefore

$\small \log (8)-\log(4)=\log(\frac{8}{4})=\log (2)$

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Question

Simplify

Answer

This problem can be solved using the properties of logs. When two logs are being subtracted from each other, it is the same thing as dividing two logs together. Remember that to use this rule, the logs must have the same base in this case .

$log_5x-log_52y=log_5\frac{x}{2y}$

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Question

Condense this logarithm: $\frac{1}{2}\log_{10}x+3\log_{10}(x+1)$

Answer

In order to solve this problem you must understand the product property of logarithms $\log_{10}uv=\log_{10}u+\log_{10}v$ and the power property of logarithms $\log_{10}u^n=n\log_{10}u$ . Note that these apply to logs of all bases not just base 10.

$\frac{1}{2}\log_{10}x+3\log_{10}(x+1)$

first move the constants in front of the logarithmic functions to their proper place using the power rule.

$\log_{10}x^{\frac{1}{2}}+\log_{10}(x+1)^3$

next factor out the logarithmic equation:

$\log_{10}[x^\frac{1}{2}+(x+1)^3]$

change the fractional exponent to a radical

$\log_{10}[\sqrt{x}(x+1)^3]$

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Question

Expand $log_{8}(\frac{15}{8})$

Answer

The rule for expanding and dividing logarithms is that you can subtract the terms inside the log. In this case, the question is not asking for an actual number, but just what the expanded version would be. Therefore you separate the terms inside the log by subtracting the denominator from the numerator. Therefore the answer is

$log_{8}(15)-log_{8}(8)$

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Question

Add the logarithms:

$log_{10}(2)+log_{10}(7)$

Answer

When adding logarithms of the same base, all you have to do is multiply the numbers inside the function as shown below:

$log_{10}(2)+log_{10}(7)$

$log_{10}(2*7)$

$log_{10}(14)$

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Question

Add the logarithms:

$log_{10}(12)+log_{10}(3)$

Answer

When adding logarithms of the same base, all you have to do is multiply the numbers inside the function as shown below:

$log_{10}(12)+log_{10}(3)$

$log_{10}(12*3)$

$log_{10}(36)$

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Question

Add the logarithms:

$log_{10}(5)+log_{10}(4)$

Answer

When adding logarithms of the same base, all you have to do is multiply the numbers inside the function as shown below:

$log_{10}(5)+log_{10}(4)$

$log_{10}(5*4)$

$log_{10}(20)$

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Question

Subtract the logarithms:

$log_{10}(14)-log_{10}(2)$

Answer

When subtracting logarithms of the same base, all you have to do is divide the numbers inside the function as shown below:

$log_{10}(14)-log_{10}(2)$

$log_{10}(14\div 2)$

$log_{10}(7)$

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Question

Subtract the logarithms:

$log_{10}(25)-log_{10}(5)$

Answer

When subtracting logarithms of the same base, all you have to do is divide the numbers inside the function as shown below:

$log_{10}(25)-log_{10}(5)$

$log_{10}(25\div 5)$

$log_{10}(5)$

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Question

Subtract the logarithms:

$log_{10}(22)-log_{10}(2)$

Answer

When subtracting logarithms of the same base, all you have to do is divide the numbers inside the function as shown below:

$log_{10}(22)-log_{10}(2)$

$log_{10}(22\div 2)$

$log_{10}(11)$

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Question

Subtract the logarithms:

$log_{10}(7)-log_{10}(4)$

Answer

When subtracting logarithms of the same base, all you have to do is divide the numbers inside the function as shown below:

$log_{10}(7)-log_{10}(4)$

$log_{10}(7\div 4)$

$log_{10}(7/4)$

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Question

Subtract the logarithms:

$log_{10}(3)-log_{10}(11)$

Answer

When subtracting logarithms of the same base, all you have to do is divide the numbers inside the function as shown below:

$log_{10}(3)-log_{10}(11)$

$log_{10}(3\div 11)$

$log_{10}(3/11)$

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Question

Subtract the logarithms:

$log_{10}(2)-log_{10}(5)$

Answer

When subtracting logarithms of the same base, all you have to do is divide the numbers inside the function as shown below:

$log_{10}(2)-log_{10}(5)$

$log_{10}(2\div 5)$

$log_{10}(2/5)$

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Question

Add the logarithms:

$log_{10}(5)+log_{10}(3)$

Answer

When adding logarithms of the same base, all you have to do is multiply the numbers inside the function as shown below:

$log_{10}(5)+log_{10}(3)$

$log_{10}(5*3)$

$log_{10}(15)$

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