Logarithms - High School Math

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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

Which of the following represents a simplified form of ?

Answer

The rule for the addition of logarithms is as follows:

.

As an application of this,.

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Question

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

Answer

Using properties of logs we get:

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

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

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Question

Simplify the following expression:

Answer

Recall the log rule:

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

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Question

Use the properties of logarithms to solve the following equation:

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

Answer

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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Question

Solve the equation.

$5^{3x-9}=25^{^{x+5}$

Answer

First, change 25 to so that both sides have the same base. Once they have the same base, you can apply log to both sides so that you can set their exponents equal to each other, which yields .

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Question

Solve the equation.

$7^{3x+4}=49^{x+1}$

Answer

Change 49 to so that both sides have the same base so that you can apply log. Then, you can set the exponential expressions equal to each other .

Thus,

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Question

Solve the equation.

$3^{3x-7}=81^{12-3x}$

Answer

Change 81 to so that both sides have the same base. Once you have the same base, apply log to both sides so that you can set the exponential expressions equal to each other (). Thus, $x=\frac{11}{3}$ .

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Question

Solve the equation.

$10^{2x-10}=(\frac{1}{100})^{8x-1}$

Answer

Change the right side to $10^{-2}$ so that both sides have the same bsae of 10. Apply log and then set the exponential expressions equal to each other

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Question

Solve the equation.

$4^{2x-7}=64^{3x}$

Answer

Change 64 to so that both sides have the same base. Apply log to both sides so that you can set the exponential expressions equal to each other

.

Thus, .

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Question

Solve the equation.

$8^{x-1}=(\frac{1}{16})^{4x-3}$

Answer

Change the left side to and the right side to $2^{-4}$ so that both sides have the same base. Apply log and then set the exponential expressions equal to each other (). Thus, $x=\frac{15}{19}$ .

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Question

Solve the equation.

$5^{11-3x}=125^{5-x}$

Answer

Change 125 to so that both sides have the same base. Apply log and then set the exponential expressions equal to each other so that . Upon trying to isolate , it becomes clear that there is no solution.

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Question

Solve the equation.

$2^{9x-3}=\left(\frac{1}{32}\right)^{x-11}$

Answer

Change the right side to $2^{-5}$ so that both sides are the same. Apply log to both sides so that you can set the exponential expressions equal to each other ().

$x=\frac{29}{7}$

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Question

Solve the equation.

$36^{10x+2}=(\frac{1}{6})^{13-x}$

Answer

Change the left side to and the right side to $6^{-1}$ so that both sides have the same base. Apply log to both sides and then set the exponential expressions equal to each other (). $x=-\frac{17}{19}$ .

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Question

Solve the equation.

$25^{20x+4}=(\frac{1}{125})^{4-3x}$

Answer

Change the left side to and the right side to $5^{-3}$ so that both sides have the same base. Apply log and then set the exponential expressions equal to each other (). Thus, $x=-\frac{20}{31}$

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Question

Solve the equation.

$27^{4x-1}=(\frac{1}{3})^{3x+8}$

Answer

Change the left side to and the right side to $3^{-1}$ so that both sides have the same base. Apply log to both sides and then set the exponential expressions equal to each other (). Thus, $x=-\frac{1}{3}$ .

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Question

Solve for .

$17 + x = \log_{10}10000$

Answer

$log_{10}10000$ can be simplified to since . This gives the equation:

Subtracting from both sides of the equation gives the value for .

$(17 + x) - 17 = (4) - 17 \rightarrow x = -13$

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Question

Solve for :

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

Answer

Since $3^{2} = 9$ , $2 = \log_{3}9$ , and we can rewrite and solve this statement as follows:

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

$\log_{3}9 + \log_{3} (x - 2) = \log_{3} (x +6)$

$\log_{3} 9 (x - 2) = \log_{3} (x +6)$

$3 ^ { \log_{3} 9 (x - 2)} =3 ^ { \log_{3} (x +6)}$

Substitution confirms this to be the only solution.

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Question

Solve for :

$\log_{4} (x-3) = \log_{16} (x-1)$

Answer

We can rewrite this as follows:

$\log_{4} (x-3) = \log_{16} (x-1)$

$\log_{4} (x-3) =\frac{ \log_{4} (x-1)}{ \log_{4} 16}$

$\log_{4} (x-3) =\frac{ \log_{4} (x-1)}{ 2}$

$2 \log_{4} (x-3) =\log_{4} (x-1)$

$\log_{4} \left [(x-3)^2\right ] =\log_{4} (x-1)$

$\dpi{120} 4^ {\log_{4} \left [(x-3)^2\right ]} =4^ { \log_{4} (x-1)}$

Split this into a compound statement, and solve:

$x - 2 \Rightarrow x = 2$

$x - 5 \Rightarrow x = 5$

Test both of these possible solutions by substituting:

$\log_{4} (x-3) = \log_{16} (x-1)$

$\log_{4} (2-3) = \log_{16} (2-1)$

$\log_{4} (-1) = \log_{16} 1$

Since only positive numbers can have logarithms, this equation does not make sense. is not a solution.

$\log_{4} (x-3) = \log_{16} (x-1)$

$\log_{4} (5-3) = \log_{16} (5-1)$

$\log_{4} 2 = \log_{16} 4$

$\frac{1}{2} =\frac{1}{2}$

is a solution - the only solution.

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Question

What is the sum of the value(s) of x that satisfy the equation $\log_{10}(2x^2-35x)=2$ ?

Answer

$\log_{10}(2x^2-35x)=2$

We need to rewrite the logarithmic equation into a form that is more recognizable.

An equation in the form $\log_{a}b=c$ can be rewritten in the exponential form , where a, b, and c are constants (and b > 0). Thus, we will rewrite our original equation as follows:

.

We can now approach this as we would a typical quadratic equation.

Subtract 100 from both sides. We need to set all of the terms equal to zero.

.

We can factor this by first multiplying the outer coefficients. The product of 2 and -100 is equal to -200. We must think of two numbers that multiply to give -200 but add to give -35 (the middle coefficient). Two numbers that satisfy both of these requirements are -40 and 5. We will now rewrite the quadratic polynomial.

We can use grouping to factor the polynomial. Factor the first two terms and the last two terms separately.

Notice that we can now factor out x - 20 from the 2x term and the 5 term.

To solve this, we set each factor equal to zero.

$x=\frac{-5}{2}$

The question ultimately asks for the sum of the values of x that satisfy the equation, so we must add -5/2 and 20, which yields 35/2.

The answer is 35/2.

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