lesson 10 practice problems
1. a. use the base - 2 log table (printed in the lesson) to approximate the value of each exponential expression.
i. $2^5$
ii. $2^{3.7}$
iii. $2^{4.25}$
b. use the base - 2 log table to find or approximate the value of each logarithm.
i. $log_2 4$
ii. $log_2 17$
iii. $log_2 35$
2. here is a logarithmic expression: $log_2 64$.
a. how do we say the expression in words?
b. explain in your own words what the expression means.
c. what is the value of this expression?
3. a. what is $log_{10}(100)$? what about $log_{100}(10)$?
b. what is $log_2(4)$? what about $log_4(2)$?
c. express $b$ as a power of $a$ if $a^2 = b$.

Question

lesson 10 practice problems
1. a. use the base - 2 log table (printed in the lesson) to approximate the value of each exponential expression.
   i. $2^5$
   ii. $2^{3.7}$
   iii. $2^{4.25}$
   b. use the base - 2 log table to find or approximate the value of each logarithm.
   i. $log_2 4$
   ii. $log_2 17$
   iii. $log_2 35$
2. here is a logarithmic expression: $log_2 64$.
   a. how do we say the expression in words?
   b. explain in your own words what the expression means.
   c. what is the value of this expression?
3. a. what is $log_{10}(100)$? what about $log_{100}(10)$?
   b. what is $log_2(4)$? what about $log_4(2)$?
   c. express $b$ as a power of $a$ if $a^2 = b$.

Accepted Answer

### Problem 1a i: Approximate $2^5$
# Explanation:
## Step1: Recall exponentiation
$2^5$ means multiplying 2 by itself 5 times.
$2^5 = 2\times2\times2\times2\times2$
## Step2: Calculate the product
$2\times2 = 4$, $4\times2 = 8$, $8\times2 = 16$, $16\times2 = 32$
# Answer: 32

### Problem 1a ii: Approximate $2^{3.7}$
(Note: Since a base - 2 log table is used, we know that if $y = 2^x$, then $x=\log_2y$. We can also use the property of exponents. $2^{3.7}=2^{3 + 0.7}=2^3\times2^{0.7}$. We know that $2^3 = 8$. To find $2^{0.7}$, from the base - 2 log table, if we consider $x = 0.7$, we can find the approximate value of $2^{0.7}$. Looking up in the base - 2 log table (assuming the table has values for fractional exponents), $2^{0.7}\approx1.6245$. Then $2^{3.7}=8\times1.6245 = 12.996\approx13$)
# Explanation:
## Step1: Split the exponent
$2^{3.7}=2^{3+0.7}$
## Step2: Use exponent rule $a^{m + n}=a^m\times a^n$
$2^{3+0.7}=2^3\times2^{0.7}$
## Step3: Calculate $2^3$
$2^3 = 8$
## Step4: Approximate $2^{0.7}$ from log table
From base - 2 log table, $2^{0.7}\approx1.6245$
## Step5: Multiply the two results
$8\times1.6245 = 12.996\approx13$
# Answer: Approximately 13

### Problem 1a iii: Approximate $2^{4.25}$
# Explanation:
## Step1: Split the exponent
$2^{4.25}=2^{4+0.25}=2^4\times2^{0.25}$
## Step2: Calculate $2^4$
$2^4=16$
## Step3: Simplify $2^{0.25}$
$2^{0.25}=2^{\frac{1}{4}}=\sqrt[4]{2}\approx1.1892$
## Step4: Multiply the two results
$16\times1.1892 = 19.0272\approx19$
# Answer: Approximately 19

### Problem 1b i: Find $\log_24$
# Explanation:
## Step1: Recall the definition of logarithm
If $\log_bx=y$, then $b^y = x$. For $\log_24$, let $y=\log_24$, then $2^y = 4$
## Step2: Solve for y
We know that $2^2 = 4$, so $y = 2$
# Answer: 2

### Problem 1b ii: Approximate $\log_217$
# Explanation:
## Step1: Recall the definition of logarithm
Let $y=\log_217$, then $2^y=17$
## Step2: Use the base - 2 log table or trial and error
We know that $2^4 = 16$ and $2^5=32$. Since $16\lt17\lt32$, $y$ is between 4 and 5. From the base - 2 log table, we can find that $2^{4.09}\approx17$ (by looking at the table values for $x$ such that $2^x = 17$)
# Answer: Approximately 4.09

### Problem 1b iii: Approximate $\log_235$
# Explanation:
## Step1: Recall the definition of logarithm
Let $y = \log_235$, then $2^y=35$
## Step2: Use the base - 2 log table or trial and error
We know that $2^5 = 32$ and $2^6=64$. Since $32\lt35\lt64$, $y$ is between 5 and 6. From the base - 2 log table, we find that $2^{5.13}\approx35$ (by looking at the table values for $x$ such that $2^x = 35$)
# Answer: Approximately 5.13

### Problem 2a: Say $\log_264$ in words
# Brief Explanations:
The expression $\log_264$ is read as "log base 2 of 64".
# Answer: log base 2 of 64

### Problem 2b: Explain the meaning of $\log_264$
# Brief Explanations:
The expression $\log_264$ represents the exponent to which we must raise the base 2 to get the number 64. In other words, if we let $y=\log_264$, we are looking for the value of $y$ such that $2^y = 64$.
# Answer: It is the exponent to which 2 must be raised to get 64.

### Problem 2c: Find the value of $\log_264$
# Explanation:
## Step1: Recall the definition of logarithm
Let $y=\log_264$, then $2^y = 64$
## Step2: Solve for y
We know that $2^6=64$ (since $2\times2\times2\times2\times2\times2 = 64$), so $y = 6$
# Answer: 6

### Problem 3a i: Find $\log_{10}100$
# Explanation:
## Step1: Recall the definition of logarithm
Let $y=\log_{10}100$, then $10^y = 100$
## Step2: Solve for y
We know that $10^2=100$ (since $10\times10 = 100$), so $y = 2$
# Answer: 2

### Problem 3a ii: Find $\log_{100}10$
# Explanation:
## Step1: Recall the definition of logarithm
Let $y=\log_{100}10$, then $100^y = 10$
## Step2: Rewrite 100 as a power of 10
$100 = 10^2$, so $(10^2)^y=10$
## Step3: Use the exponent rule $(a^m)^n=a^{mn}$
$10^{2y}=10^1$
## Step4: Equate the exponents
$2y = 1$, so $y=\frac{1}{2}=0.5$
# Answer: 0.5

### Problem 3b i: Find $\log_24$
# Explanation:
## Step1: Recall the definition of logarithm
Let $y=\log_24$, then $2^y = 4$
## Step2: Solve for y
We know that $2^2 = 4$, so $y = 2$
# Answer: 2

### Problem 3b ii: Find $\log_42$
# Explanation:
## Step1: Recall the definition of logarithm
Let $y=\log_42$, then $4^y = 2$
## Step2: Rewrite 4 as a power of 2
$4 = 2^2$, so $(2^2)^y=2$
## Step3: Use the exponent rule $(a^m)^n=a^{mn}$
$2^{2y}=2^1$
## Step4: Equate the exponents
$2y = 1$, so $y=\frac{1}{2}=0.5$
# Answer: 0.5

### Problem 3c: Express $b$ as a power of $a$ if $a^2 = b$
# Explanation:
## Step1: Recall the definition of exponentiation
If $a^2=b$, by the definition of exponentiation, $b$ is $a$ raised to the power of 2.
# Answer: $b = a^2$

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QUESTION IMAGE

Question

Explanation:

Problem 1a i: Approximate \(2^5\)

Step1: Recall exponentiation

Step2: Calculate the product

Step1: Split the exponent

Step2: Use exponent rule \(a^{m + n}=a^m\times a^n\)

Step3: Calculate \(2^3\)

Step4: Approximate \(2^{0.7}\) from log table

Step5: Multiply the two results

Step1: Split the exponent

Step2: Calculate \(2^4\)

Step3: Simplify \(2^{0.25}\)

Step4: Multiply the two results

Answer:

Problem 1a ii: Approximate \(2^{3.7}\)