Multiplying and dividing Logarithms

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Math › Multiplying and dividing Logarithms

Questions 1 - 5

1

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.

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

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

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

2

Which of the following represents a simplified form of ?

$\frac{log(5) }{ log(x) }$

Explanation

The rule for the addition of logarithms is as follows:

.

As an application of this,.

3

Simplify $log_{5}75 - log_{5}3$ .

Explanation

Using properties of logs we get:

$log_{5}75 - log_{5}3 = log_{5}\frac{75}{3}= log_{5}25$

$log_{5}25= log_{5}5^{2}= 2log_{5}5= 2\cdot 1=2$

4

Simplify the following expression:

Explanation

Recall the log rule:

In this particular case, and . Thus, our answer is $log(4\cdot 12) = log(48)$ .

5

Use the properties of logarithms to solve the following equation:

$log_{6}(n-3)+log_{6}(n+2)=log_{3}(3)$

No real solutions

Explanation

Since the bases of the logs are the same and the logarithms are added, the arguments can be multiplied together. We then simplify the right side of the equation:

$log_{6}(n-3)+log_{6}(n+2)=log_{3}(3)$

$log_{6}(n-3)(n+2)=1$

The logarithm can be converted to exponential form:

$(n-3)(n+2)=6^{1}$

Factor the equation:

Although there are two solutions to the equation, logarithms cannot be negative. Therefore, the only real solution is .

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