Logarithms

Question 1

What are the components of logarithm?

Answer 1

The“log”function, which states that we are dealing with a logarithmThebaseor the small number below the “log” symbolTheargumentor the larger number on the right of the base.Finally, the number on the right of the equal sign is theexponent.

Question 2

Let us compute for log416. This means we must think of the exponent used for 4 to get a value of 16.

Answer 2

Note that if we multiply 4 by itself, we can get 16 (i.e., 4 x 4 = 16); hence, we must use an exponent of 2 for 4 to get 16.

Therefore, log416 = 2.

Question 3

Sample Problem:Compute the value of the following:

log327
log232
log525
log71

Answer 3

Solution:

Note that if we multiply 3 by itself three times, we can get 27 (i.e., 3 x 3 x 3 = 27). Therefore, log327 = 3.If we multiply 2 by itself five times, we can get 32 ( 2 x 2 x 2 x 2 x 2 = 32). Therefore, log232 = 5
.Multiplying 5 by itself two times will result in 25. Therefore, log525 = 2.To evaluate the value of log71, we think of a power to which we must raise 7 so that we obtain 1. By the zero-exponent rule, we know that any real number raised to zero is equal to 1. Hence, we must raise 7 to the power of 0 to get 1. Therefore, log71 = 0.

Question 4

log6 1 tells us what exponent we should raise 6 so that we can get 1.

Answer 4

Recall that thezero-exponent rulestates that any real number raised to 0 will result in 1. So, if we raise 6 to the power of 0, we can obtain 1. Mathematically, 60= 1.

Therefore, the value of log61 must be equal to 0 (log61 = 0).

To sum up,if the argument of the logarithm is 1, then the value of the logarithm is automatically 0 as per the zero-exponent rule:

loga1 = 0, where a is a real number

Question 5

To convert a quantity in exponential form into logarithmic form, follow the steps below:

Answer 5

Write the “log” sign to indicate that you’re using the logarithm operatorWrite the base of the exponential form as the base of the logarithmic form (i.e., the small number on the right of the log)
Write the value of the exponential form as the argument of the logarithmic form (i.e., the number on the right of the base of the logarithmic form)
Write the exponential form as the exponent of the logarithmic form

In general, a quantity in exponential form by= x is written as logbx = y in logarithmic form.

Question 6

Sample Problem:Write 53= 125 into logarithmic form.

Answer 6

Solution:Using the steps on transforming the exponential form into the logarithmic form:

Step 1:Write the “log” sign to indicate that you’re using the logarithm operator

Note: the boxes in the expression above serve as placeholders for the remaining components of the logarithm.

Step 2:Write the base of the exponential form as the base of the logarithmic form (i.e., the small number on the right of the log)

The base of 53= 125 is 5. Hence we use it as the base of the logarithmic form:

Step 3:Write the value of the exponential form as the argument of the logarithmic form (i.e., the number on the right of the base of the logarithmic form)

The value of the exponential form 53= 125 is 125. Hence, we will use it as the argument of the logarithmic form.

Step 4:Write the exponential form as the exponent of the logarithmic form

Finally, we use “3” (the exponent in 53= 125) as the exponent in the logarithmic form.

log5125 = 3

Question 7

To transform a quantity expressed in logarithmic form into exponential form, we have to follow these steps:

Answer 7

Drop the “log” signUse the base of the logarithmic form (i.e., the small number on the right of the “log” symbol) as the base of the exponential formUse the exponent of the logarithmic form (i.e., the number on the right of the equal sign) as the exponent for the exponential formUse the argument as the value of the exponential form

Question 8

Sample Problem:Express log5125 = 3 into exponential form

Answer 8

Solution:

Step 1:Drop the “log” sign

Step 2:Use the base of the logarithmic form (i.e., the small number on the right of the “log” symbol) as the base of the exponential form

The base of the logarithmic form is 5, so we use it as the base of the exponential form:

Step 3:Use the exponent of the logarithmic form (i.e., the number on the right of the equal sign) as the exponent for the exponential form

The exponent of the logarithmic form is 3 so we’ll use it as the exponent in the exponential form.

Step 4:Use the argument as the value of the exponential form

Finally, we use 125 (the argument in log5125 = 3) as the value of the exponential form.

Therefore,log5125 = 3 can be written as 5^3= 125.

Question 9

In a ____ logarithm, we don’t need to write the “10” as the base anymore. We just leave it blank since it is already understood that we are dealing with a common logarithm.

Answer 9

Common log

Question 10

Sample Problem:Evaluate log 1000

Answer 10

Solution:log 1000 is a common logarithm. Although it appears that it has no base, it means that the base is 10. Therefore, log 1000 = log101000.

To find the value of log101000, we need to think of how many times 10 must be multiplied by itself to get 1000. Note that if we multiply 10 by itself thrice, the result will be 1000 (10 x 10 x 10 = 1000). Thus, log101000 = 3.

Question 11

_____ logarithms use a “special number” as the base. This number is anirrational numberrepresented ase. Formally,eis called Euler’s number (named after the mathematician Leonhard Euler).eis approximately equal to 2.718…

Answer 11

Natural log

Question 12

This property states that the logarithm of products can be expressed as the sum of the logarithms:

logaPQ = logaP + logaQ

Answer 12

Product Property of Logarithms

Question 13

Sample Problem 1:Express log25 + log24 as a single logarithm.

Answer 13

Solution:By applying the product property of logarithms, we just simply multiply the arguments of the given logarithms.

log25 + log24 = log2(5 x 4) = log2 20

Question 14

Sample Problem 2: Given thatlog 3 ≈ 0.48 and log 2 ≈ 0.30. What is the approximate value of log 6?

Answer 14

Solution:Since log 3 and log 2 are both common logarithms, they have the same base which is 10. This means that we can apply the product property to them.

Take note that “log 3 ≈ 0.48” means that the approximate value of log 3 is 0.48. This means that the value of log 3 is not exactly 0.48 since this value is just an approximation. The “ ≈ “ symbol denotes an approximation of a particular quantity.

Applying the product property, we have: log 6 = log (3 x 2) = log3 + log 2

This means that the value of log 6 is equal to the value of log 3 + log 2:

log 6 = log 3 + log 2

Using the given approximate values of log 3 and log 2:

log 6 = log 3 + log 2

log 6 ≈ 0.48 + 0.30

log 6 ≈ 0.78

Hence, the approximate value of log 6 is 0.78 (the value of log 6 when computed using a calculator is 0.77815… which is extremely near to our obtained value).

Question 15

Sample Problem 3:Expand log4(7a(b + 4)) using the product property of logarithms.

Answer 15

Solution:Note that in log4(7a(b + 4)), the argument 7a(b + 4) is the product of 7a and b + 4. Furthermore, 7a is also the product of two quantities: 7 and a.

Therefore, log4(7a(b + 4)) = log4(7 ∙ a ∙ (b + 4)). By product rule, we can now express this as the sum of logarithms:

log4(7a(b + 4)) = log4(7 ∙ a ∙ (b + 4)) = log47 + log4a + log4(b + 4)

Question 16

This property states that the logarithm of the quotient can be expressed as the difference of logarithms.

Question 17

Expand log (x-1/3)

Answer 17

Solution:The argument of is

which is the quotient of x – 1 and 3. By the quotient property of logarithms, we can express the logarithm of the quotient of x – 1 and 3 as the difference between the logarithm of x – 1 and the logarithm of 3.

By applying the quotient property of logarithms:

= log (x-1) - log 3

Question 18

Expand log (9/b)

Answer 18

Solution:The argument is the quotient of 9 and b. By the quotient property of logarithms, we can express the logarithm of a quotient of 9 and b as the difference between the logarithm of 9 and the logarithm of b.

By applying the quotient property of logarithms:

log 9 - log b

Question 19

states that the logarithm of a quantity raised to an exponent is equal to the exponent times the logarithm of the quantity.

logaPq= q∙logaP

Answer 19

Power Property of Logarithms

Question 20

Sample Problem 1:Expand loga^3

Answer 20

Solution:Since we are dealing with a logarithm that has an argument with a quantity raised to an exponent (a3), we can apply the power property of logarithms.

log a3= 3 ∙ log a = 3 log a

Hence, the answer is 3 log a.

Question 21

Sample Problem 2:Expand ln j2

Answer 21

Solution:By applying the product property of logarithms:

ln j2= 2 ∙ ln j = 2 ln j

Therefore, the answer is 2 ln j.

Question 22

Sample Problem 1:Expand log4a^2b

Answer 22

Solution:The argument in the expression log4a2b is a2b which is the product of a2and b. Hence, we can apply theproduct property of logarithmsto start expanding the given expression:

log4a2b = log4a2+ log4b

Note that we can still expand the expression log4a2since we have an argument with a quantity raised to an exponent (a2). We can apply thepower property of logarithmsfor this case:

log4a2b = log4a2+ log4b

log4a2b = 2 log4a + log4b

Hence, the expanded form of log4a2b is 2 log4a + log4b.

Here’s the summary of what we have performed above:

log4a2b

log4a2+ log4bProduct Property of Logarithms

2 log4a + log4bPower Property of Logarithms

Question 23

Sample Problem 2:Expand ln x^2y^3

Answer 23

Solution:The argument in the expression ln x2y3is x2y3or x2and y3. This means we can apply theproduct property of logarithmsto expand the given expression.

ln x2y3= ln x2+ ln y3

Note that we can still expand both ln x2and ln y3since they involve arguments that are quantities raised to an exponent (x2and y3). Thus, we can apply thepower property of logarithmsfor this case:

ln x2y3= ln x2+ ln y3

ln x2y3= 2 ln x + 3 ln y

Hence, the answer for this example is 2 ln x + 3 ln y.

Here’s the summary of our calculation above:

ln x2y3

ln x2+ ln y3Product Property of Logarithms

2 ln x + 3 ln yPower Property of Logarithms

Question 24

Sample Problem 3:

Expand log (3a^2/2(a+b))

Answer 24

The argument of the given logarithm is the quotient of 3a2and 2(a + b). Hence, we can apply the quotient property to expand the given logarithm.

Notice that we can expand both log 3a2and log 2(a + b). Let us expand log 3a2first.

The argument of log 3a2is 3a2which is the product between 3 and a2. By product property:

log 3a2= log 3 + log a2

We can still expand log a2by applying the power property:

log 3a2= log 3 + 2 log a

Therefore, we now have:

Now, let us focus on log 2(a + b). The argument here is 2(a + b) which is the product of 2 and a + b. Hence, we can apply the product property:

Thus, the final answer is
(log 3 + 2 log a) – ( log 2 + log(a + b))

Question 25

express log310 + log35 as a single logarithm.

Answer 25

The expression log310 + log35 involves themathematical operation of addition. Note that we have learned in the previous section that the logarithm of a product can be expressed as the sum of logarithms (product property). Hence, we can apply the reverse of this property and express the sum of logarithms as the logarithms of a product.

By product property, we can express log310 + log35 as the logarithm of the product of 10 and 5 (to the base 3):

log310 + log35

log3(10 x 5) = log350Product Property of Logarithms

Hence, the simplified form of log310 + log35 is log350.

Question 26

Sample Problem 1:Simplify log28 + log27

Answer 26

Solution:By the product property, we can express the sum of logarithms as the logarithm of their product:

log28 + log27

log2(8 x 7)Product Property of Logarithms

log256

Thus, the answer is log256.

Question 27

Sample Problem 2:Simplify log3(x + 5) + log3x + log34

Answer 27

Solution:By the product property of logarithms, we can express the sum of logarithms as the logarithm of their product:

log3(x + 5) + log3x + log34

log3[(x + 5) ∙ x ∙ 4]

log3(4(x)(x + 5))

log3(4x(x + 5))

log3(4x2+ 20x)

Thus, the simplified form is log3(4x^2+ 20x).

Question 28

Sample Problem 3:Apply the properties of logarithms to simplify ln 6 + ln 3 – ln 2

Answer 28

Solution:We are now dealing with two operations in the given logarithmic expression: addition and subtraction.

Let us deal with the addition sign first. By product property, we can express the sum of logarithms as the logarithm of their product:

ln 6 + ln 3 – ln 2

ln (6 x 3) – ln 2

ln 18 – ln 2

The resulting logarithmic expression ln 18 – ln 2 can be simplified further. The difference between logarithms can be expressed as the logarithm of their quotient using the quotient property. Thus:

ln 18 – ln 2 = ln (18∕2) = ln 9

Thus, the final answer is ln 9.

Question 29

Sample Problem 4:Simplify 3 log2a + log2b

Answer 29

Solution:Note that by power rule, we can express the product of a constant and the logarithm of a quantity as the logarithm of the quantity raised to the constant. Note that 3 log2a is the product of a constant (which is 3) and a logarithm of a quantity (log2a). Hence, we can apply the power rule to simplify it:

3 log2a + log2b

log2a3+ log2bPower Rule of Logarithms

Now, we can simply apply the product rule to complete the solution:

log2a3+ log2b

log2a3bProduct Rule of Logarithms

Thus, the final answer is log2 a^3b.

Question 30

1) Express log5 x + log5 y = z in exponential form

(a) 5^z = x + y
(b) 5^x = zy
(c) 5^z = xy
(d) 5^x = z + y

Answer 30

C

Question 31

2) What is the value of 1-log 100/log 100+1?

(a) -⅓
(b) ½
(c) -½
(d) ⅓

Answer 31

A

Question 32

3) Expand loga(2(x - y))^b

(a) b loga2 + loga(x - y)
(b) b (loga2 + loga(x - y))
(c) b loga2x + logy
(d) loga2 + b loga(x - y)

Answer 32

B

Question 33

4) If logap = 3 and logaq = 4, what
is the value of (loga p/q)^2
(a) -2
(b) 2
(c) 1
(d) -1

Answer 33

C

Question 34

5) Express 9 log2x + 3 log2y as a single logarithm.

(a) log2 9𝑥/3𝑦
(b) log2 𝑥^9𝑦^3
(c) log2 9𝑥𝑦^3
(d) None of the above

Answer 34

B