Algebra II - Adding and Subtracting Logarithms

Simplify the expression using logarithmic identities.

$log_4(\frac{7}{5})$

$log_4(\frac{5}{7})$

The expression cannot be simplified

Explanation

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.

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

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

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

Which is another way of expressing

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

$\small \log(20)$

$\small \log(\frac{4}{5})$

$\small \log(9)$

$\small \small \frac{\log (4)}{\log (5)}$

$\small \log _{4}(5)$

Explanation

Use the rule:

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

therefore

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

Add the logarithms:

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

$log_{10}(14)$

$log_{10}(9)$

$log_{10}(4)$

$log_{10}(12)$

Explanation

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

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

$log_{10}(2*7)$

$log_{10}(14)$

Combine as one log:

Explanation

According to log rules, the coefficients of the logs can be raised as powers for the inner quantity of the log. Rewrite the terms.

Subtract the .

The answer is:

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

False

True

Explanation

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

$10 ^{1} = e$ ,

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

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

True

False

Explanation

By the Change of Base Property of Logarithms, if and ,

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

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."

True or false: for all positive values of ,

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

True

False

Explanation

By the Product of Logarithms Property,

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

Setting , 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

$10^{a} = 100$

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

so $\log 100 = 2$

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

The statement is true.

True or false:

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

for all negative values of .

False

True

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.

Add the logarithms:

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

$log_{10}(20)$

$log_{10}(2)$

$log_{10}(9)$

$log_{10}(12)$

Explanation

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

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

$log_{10}(5*4)$

$log_{10}(20)$

Simplify the following expression:

$\log_{3}6+\log_{3}4-\log_{4}5$

$\log_324-\log_45$

$\log_324+\log_45$

$\log_310-\log_45$

$\log_310+\log_45$

Explanation

The furthest we can simplify the given expression is combining the two terms with the same base by multiplying the numbers on the "outside" of the logarithm (addition becomes multiplication when performing logarithmic operations).

Our final answer is therefore

$\log_324-\log_45$

Adding and Subtracting Logarithms

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