Properties Of Logs- Review For Quiz

1. Describe the following expression as a single logarithm: (1/2)logm + (1/3) logn

Log( cube root of m, times, n)

Log (square root of m, times, the cube root of n)

Log( 6th root of mn)

Log (square roots of m, divided by, the cube root of n)

The given expression (1/2)logm + (1/3)logn can be simplified by using the property of logarithms that states loga + logb = log(ab). Applying this property, we can rewrite the expression as log(sqrt(m)) + log(cbrt(n)). Since the logarithm of a product is equal to the sum of the logarithms of its factors, this can be further simplified to log(sqrt(m) * cbrt(n)). Therefore, the correct answer is "log (square root of m, times, the cube root of n)".

Explanation

The given expression (1/2)logm + (1/3)logn can be simplified by using the property of logarithms that states loga + logb = log(ab). Applying this property, we can rewrite the expression as log(sqrt(m)) + log(cbrt(n)). Since the logarithm of a product is equal to the sum of the logarithms of its factors, this can be further simplified to log(sqrt(m) * cbrt(n)). Therefore, the correct answer is "log (square root of m, times, the cube root of n)".

2. Write as a single logarithm: 3logm - 2logn - 0.5logp

Log (m3n2p)

Log(m/(n2p)0.5

Log (m3/(n2p0.5)

Log(1/(m3n2p)

The given expression can be simplified by using the properties of logarithms. By applying the power rule of logarithms, we can rewrite the expression as log(m^3) - log(n^2) - log(p^0.5). Then, using the quotient rule of logarithms, we can combine the terms to get log(m^3/(n^2*p^0.5)). Therefore, the correct answer is log(m^3/(n^2*p^0.5)).

Explanation

The given expression can be simplified by using the properties of logarithms. By applying the power rule of logarithms, we can rewrite the expression as log(m^3) - log(n^2) - log(p^0.5). Then, using the quotient rule of logarithms, we can combine the terms to get log(m^3/(n^2*p^0.5)). Therefore, the correct answer is log(m^3/(n^2*p^0.5)).

3. A logarithmic function is the inverse of which function?

Quadratic

Linear

Cubic

Exponential

A logarithmic function is the inverse of an exponential function. This means that if we have an exponential function y = a^x, the logarithmic function that undoes this operation is y = log base a of x. The logarithmic function "undoes" the exponential function by finding the power to which the base must be raised to obtain a certain value. Therefore, the correct answer is exponential.

Explanation

A logarithmic function is the inverse of an exponential function. This means that if we have an exponential function y = a^x, the logarithmic function that undoes this operation is y = log base a of x. The logarithmic function "undoes" the exponential function by finding the power to which the base must be raised to obtain a certain value. Therefore, the correct answer is exponential.

4. What is the parent function of : f(x)=log₂(x-1)+3

F(x)=log2x

F(x)=log2(x-1)

F(x)=2x

F(x)=ln(2)

The correct answer is f(x)=log2x. This is because the given function f(x)=log2(x-1)+3 is a transformation of the parent function f(x)=log2x. The +3 in the given function represents a vertical shift upwards by 3 units, but the base function remains the same. Therefore, the parent function is f(x)=log2x.

Explanation

The correct answer is f(x)=log2x. This is because the given function f(x)=log2(x-1)+3 is a transformation of the parent function f(x)=log2x. The +3 in the given function represents a vertical shift upwards by 3 units, but the base function remains the same. Therefore, the parent function is f(x)=log2x.

5. Logarithmic functions always have _____ asymptotes?

Logarithmic functions always have vertical asymptotes. This is because the domain of a logarithmic function is restricted to positive real numbers. As the input approaches zero, the output of the logarithmic function approaches negative infinity. Therefore, the graph of a logarithmic function approaches a vertical line, which is the vertical asymptote, as the input approaches zero.

Explanation

Logarithmic functions always have vertical asymptotes. This is because the domain of a logarithmic function is restricted to positive real numbers. As the input approaches zero, the output of the logarithmic function approaches negative infinity. Therefore, the graph of a logarithmic function approaches a vertical line, which is the vertical asymptote, as the input approaches zero.

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6. The equation 2 ^x = 9 can be entered in the calculator in which form?

(log 9) / (log 2)

Log 9 * log 2

(log 9 )^2

(log 2) / (log 9)

To solve the equation 2x = 9, we need to isolate x. One way to do this is by using logarithms. Taking the logarithm of both sides of the equation, we get log(2x) = log(9). By the logarithmic property, we can bring down the exponent, so x * log(2) = log(9). To isolate x, we divide both sides of the equation by log(2), giving us x = log(9) / log(2). Therefore, the correct form to enter the equation 2x = 9 into the calculator is (log 9) / (log 2).

Explanation

To solve the equation 2x = 9, we need to isolate x. One way to do this is by using logarithms. Taking the logarithm of both sides of the equation, we get log(2x) = log(9). By the logarithmic property, we can bring down the exponent, so x * log(2) = log(9). To isolate x, we divide both sides of the equation by log(2), giving us x = log(9) / log(2). Therefore, the correct form to enter the equation 2x = 9 into the calculator is (log 9) / (log 2).

7. The following expression: 3 - log₃n is the correct expansion of : log₃(27/n)

True

False

The expression 3 - log3n is the correct expansion of log3(27/n) because when we apply the logarithmic property loga(b/c) = loga(b) - loga(c), we can rewrite log3(27/n) as log3(27) - log3(n), which simplifies to 3 - log3n.

Explanation

The expression 3 - log3n is the correct expansion of log3(27/n) because when we apply the logarithmic property loga(b/c) = loga(b) - loga(c), we can rewrite log3(27/n) as log3(27) - log3(n), which simplifies to 3 - log3n.

8. Condense using log properties, then simplify «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«msub»«mi mathvariant=¨normal¨»log«/mi»«mn»4«/mn»«/msub»«mn»2«/mn»«mo»+«/mo»«msub»«mi mathvariant=¨normal¨»log«/mi»«mn»4«/mn»«/msub»«mn»32«/mn»«mo».«/mo»«/math»

«math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«msub»«mi mathvariant=¨normal¨»log«/mi»«mn»4«/mn»«/msub»«mn»2«/mn»«mo»+«/mo»«msub»«mi mathvariant=¨normal¨»log«/mi»«mn»4«/mn»«/msub»«mn»32«/mn»«mo».«/mo»«/math»

2

3

4

8

The given expression can be simplified by using the property of logarithms that states log(a) + log(b) = log(a * b). By applying this property, we can condense the expression into log(2 * 3 * 4 * 8). Further simplification gives us log(192), which is equal to 3.

Explanation

The given expression can be simplified by using the property of logarithms that states log(a) + log(b) = log(a * b). By applying this property, we can condense the expression into log(2 * 3 * 4 * 8). Further simplification gives us log(192), which is equal to 3.

9. Expand: log₂8mn

3 + log2m + log2n

3 + log2m - log2n

Log28 + log2mn

2 + log28 - log2mn

The given expression is log28mn. Using the logarithmic property log(ab) = log(a) + log(b), we can rewrite the expression as log28 + log2m + log2n. Therefore, the correct answer is 3 + log2m + log2n.

Explanation

The given expression is log28mn. Using the logarithmic property log(ab) = log(a) + log(b), we can rewrite the expression as log28 + log2m + log2n. Therefore, the correct answer is 3 + log2m + log2n.

10. Use the Change of Base formula to evaluate the following: log₃20

2.718

0.367

0

2.727

The correct answer is 2.727. To evaluate log320, we can use the change of base formula which states that log base a of b can be calculated as log base c of b divided by log base c of a. In this case, we can use the natural logarithm (log base e) as the base c. So, log320 can be calculated as log base e of 20 divided by log base e of 3. Evaluating this expression gives us approximately 2.727.

Explanation

The correct answer is 2.727. To evaluate log320, we can use the change of base formula which states that log base a of b can be calculated as log base c of b divided by log base c of a. In this case, we can use the natural logarithm (log base e) as the base c. So, log320 can be calculated as log base e of 20 divided by log base e of 3. Evaluating this expression gives us approximately 2.727.