ppc lesson 5.4 homework
name
1. evaluate:
a. $\log_{4} 64$
b. $\log_{2} 32$
c. $\log_{7} 49$
d. $\log 10000$
2. complete the table of selected values for the exponential function $f$.
$\begin{array}{c|c|c|c|c|c|c} x & 0 & & -1 & & & \\ \hline f(x)=4^{x} & & 4 & & 2 & 256 & \frac{1}{16} \\ \end{array}$
3. determine if each logarithm is defined. if not, explain why not. if it is defined, determine if the value is a whole number or not.
a. $\log_{3} 12$
b. $\log_{3} (-9)$
c. $\log_{25} 5$
d. $\log_{4} \frac{1}{64}$

Question

ppc lesson 5.4 homework
name
1. evaluate:
 a. $\log_{4} 64$
 b. $\log_{2} 32$
 c. $\log_{7} 49$
 d. $\log 10000$
2. complete the table of selected values for the exponential function $f$.
 $\begin{array}{c|c|c|c|c|c|c} x & 0 &  & -1 &  &  &  \\ \hline f(x)=4^{x} &  & 4 &  & 2 & 256 & \frac{1}{16} \\ \end{array}$
3. determine if each logarithm is defined. if not, explain why not. if it is defined, determine if the value is a whole number or not.
 a. $\log_{3} 12$
 b. $\log_{3} (-9)$
 c. $\log_{25} 5$
 d. $\log_{4} \frac{1}{64}$

Accepted Answer

### Problem 1a: Evaluate $\boldsymbol{\log_{4} 64}$
# Explanation:
## Step1: Recall the logarithm definition
If $\log_{b} a = x$, then $b^{x} = a$. We need to find $x$ such that $4^{x} = 64$.
## Step2: Express 64 as a power of 4
We know that $4^{3} = 64$ (since $4	imes4	imes4 = 64$).
## Step3: Determine the value of the logarithm
Since $4^{3}=64$, by the definition of logarithm, $\log_{4} 64 = 3$.
# Answer:
$3$

### Problem 1b: Evaluate $\boldsymbol{\log_{2} 32}$
# Explanation:
## Step1: Recall the logarithm definition
If $\log_{b} a = x$, then $b^{x} = a$. We need to find $x$ such that $2^{x} = 32$.
## Step2: Express 32 as a power of 2
We know that $2^{5}=32$ (since $2	imes2	imes2	imes2	imes2 = 32$).
## Step3: Determine the value of the logarithm
Since $2^{5} = 32$, by the definition of logarithm, $\log_{2} 32=5$.
# Answer:
$5$

### Problem 1c: Evaluate $\boldsymbol{\log_{7} 49}$
# Explanation:
## Step1: Recall the logarithm definition
If $\log_{b} a = x$, then $b^{x} = a$. We need to find $x$ such that $7^{x}=49$.
## Step2: Express 49 as a power of 7
We know that $7^{2} = 49$ (since $7	imes7 = 49$).
## Step3: Determine the value of the logarithm
Since $7^{2}=49$, by the definition of logarithm, $\log_{7} 49 = 2$.
# Answer:
$2$

### Problem 1d: Evaluate $\boldsymbol{\log 10000}$
# Explanation:
## Step1: Recall the logarithm base (implicit base 10)
The logarithm $\log a$ means $\log_{10} a$. So we need to find $x$ such that $10^{x}=10000$.
## Step2: Express 10000 as a power of 10
We know that $10^{4}=10000$ (since $10	imes10	imes10	imes10 = 10000$).
## Step3: Determine the value of the logarithm
Since $10^{4}=10000$, by the definition of logarithm, $\log 10000 = 4$.
# Answer:
$4$

### Problem 2: Complete the table for $\boldsymbol{f(x) = 4^{x}}$
The table is:
| $x$ | $0$ |  | $-1$ |  |  |  |
| --- | --- | --- | --- | --- | --- | --- |
| $f(x)=4^{x}$ |  | $4$ |  | $2$ | $256$ | $\frac{1}{16}$ |

We will find the missing values by using the function $f(x)=4^{x}$ (i.e., for a given $x$, compute $4^{x}$; for a given $f(x)$, solve $4^{x}=f(x)$ for $x$).

#### For $x = 0$:
# Explanation:
## Step1: Substitute $x = 0$ into $f(x)$
$f(0)=4^{0}$.
## Step2: Recall the exponent rule $a^{0}=1$ ($a
eq0$)
So $4^{0}=1$.
# Answer:
When $x = 0$, $f(x)=1$.

#### For $f(x)=4$:
# Explanation:
## Step1: Set up the equation
$4^{x}=4$.
## Step2: Express 4 as a power of 4
$4 = 4^{1}$. So $4^{x}=4^{1}$.
## Step3: Since the bases are equal, exponents are equal
So $x = 1$.
# Answer:
When $f(x)=4$, $x = 1$.

#### For $x=-1$:
# Explanation:
## Step1: Substitute $x=-1$ into $f(x)$
$f(-1)=4^{-1}$.
## Step2: Recall the exponent rule $a^{-n}=\frac{1}{a^{n}}$
So $4^{-1}=\frac{1}{4}$.
# Answer:
When $x=-1$, $f(x)=\frac{1}{4}$.

#### For $f(x)=2$:
# Explanation:
## Step1: Set up the equation
$4^{x}=2$.
## Step2: Express 4 as $2^{2}$
So $(2^{2})^{x}=2$. Using the exponent rule $(a^{m})^{n}=a^{mn}$, we get $2^{2x}=2^{1}$.
## Step3: Since the bases are equal, exponents are equal
So $2x = 1\implies x=\frac{1}{2}$.
# Answer:
When $f(x)=2$, $x=\frac{1}{2}$.

#### For $f(x)=256$:
# Explanation:
## Step1: Set up the equation
$4^{x}=256$.
## Step2: Express 256 as a power of 4
We know that $4^{4}=256$ (since $4	imes4	imes4	imes4 = 256$). So $4^{x}=4^{4}$.
## Step3: Since the bases are equal, exponents are equal
So $x = 4$.
# Answer:
When $f(x)=256$, $x = 4$.

#### For $f(x)=\frac{1}{16}$:
# Explanation:
## Step1: Set up the equation
$4^{x}=\frac{1}{16}$.
## Step2: Express $\frac{1}{16}$ as a power of 4
We know that $\frac{1}{16}=\frac{1}{4^{2}}=4^{-2}$ (using $a^{-n}=\frac{1}{a^{n}}$). So $4^{x}=4^{-2}$.
## Step3: Since the bases are equal, exponents are equal
So $x=-2$.
# Answer:
When $f(x)=\frac{1}{16}$, $x=-2$.

Putting it all together, the completed table is:

| $x$ | $0$ | $1$ | $-1$ | $\frac{1}{2}$ | $4$ | $-2$ |
| --- | --- | --- | --- | --- | --- | --- |
| $f(x)=4^{x}$ | $1$ | $4$ | $\frac{1}{4}$ | $2$ | $256$ | $\frac{1}{16}$ |

### Problem 3a: Determine if $\boldsymbol{\log_{3} 12}$ is defined. If defined, is the value a whole number?
# Explanation:
## Step1: Recall the definition of a logarithm
The logarithm $\log_{b} a$ is defined if $a>0$ and $b>0$, $b
eq1$. Here, $b = 3$ (which is $>0$, $b
eq1$) and $a = 12$ (which is $>0$). So $\log_{3}12$ is defined.
## Step2: Check if the value is a whole number
We know that $\log_{3}12=\frac{\log 12}{\log 3}$ (change - of - base formula). $\log 12\approx1.0792$, $\log 3\approx0.4771$, so $\log_{3}12\approx\frac{1.0792}{0.4771}\approx2.262$. Since $2.262$ is not a whole number.
# Answer:
$\log_{3}12$ is defined, and its value is not a whole number.

### Problem 3b: Determine if $\boldsymbol{\log_{3}(-9)}$ is defined. If not, explain why.
# Explanation:
## Step1: Recall the definition of a logarithm
The logarithm $\log_{b} a$ is defined only when $a>0$. Here, $a=-9$ which is less than $0$.
## Step2: Conclusion
Since the argument of the logarithm (the value inside the logarithm) is negative, $\log_{3}(-9)$ is not defined.
# Answer:
$\log_{3}(-9)$ is not defined because the argument of a logarithm (the value inside the $\log$) must be positive, and here the argument is $-9$ (which is negative).

### Problem 3c: Determine if $\boldsymbol{\log_{25} 5}$ is defined. If defined, is the value a whole number?
# Explanation:
## Step1: Check the definition
For $\log_{25}5$, $b = 25>0$, $b
eq1$, and $a = 5>0$. So it is defined.
## Step2: Evaluate the logarithm
Let $\log_{25}5=x$. Then by the definition of logarithm, $25^{x}=5$. We can write $25$ as $5^{2}$, so $(5^{2})^{x}=5$. Using the exponent rule $(a^{m})^{n}=a^{mn}$, we get $5^{2x}=5^{1}$. Since the bases are equal, $2x = 1\implies x=\frac{1}{2}$. $\frac{1}{2}$ is not a whole number.
# Answer:
$\log_{25}5$ is defined, and its value is not a whole number.

### Problem 3d: Determine if $\boldsymbol{\log_{4}\frac{1}{64}}$ is defined. If defined, is the value a whole number?
# Explanation:
## Step1: Check the definition
For $\log_{4}\frac{1}{64}$, $b = 4>0$, $b
eq1$, and $a=\frac{1}{64}>0$. So it is defined.
## Step2: Evaluate the logarithm
Let $\log_{4}\frac{1}{64}=x$. Then $4^{x}=\frac{1}{64}$. We can write $\frac{1}{64}$ as $4^{-3}$ (since $4^{3}=64$, so $\frac{1}{64}=4^{-3}$). So $4^{x}=4^{-3}$. Since the bases are equal, $x=-3$. $-3$ is an integer (and thus a whole number? Wait, whole numbers are non - negative integers (0, 1, 2, 3,...). Wait, no: Wait, in some definitions, whole numbers are 0, 1, 2, 3,... (non - negative integers), and integers include negative numbers. Wait, let's re - check.

Wait, the problem says "whole number". Let's confirm:

We have $\log_{4}\frac{1}{64}=x\implies4^{x}=\frac{1}{64}$. Since $4^{3}=64$, $\frac{1}{64}=4^{-3}$, so $x=-3$. Now, whole numbers are typically defined as 0, 1, 2, 3,.... So $-3$ is an integer but not a whole number? Wait, no, maybe the problem considers integers (including negative) as whole numbers? Wait, no, standard definition: Whole numbers are the set $\{0, 1, 2, 3,\dots\}$, integers are $\{\dots, - 3,-2,-1,0,1,2,3,\dots\}$.

Wait, let's re - evaluate:

$4^{x}=\frac{1}{64}$. $64 = 4^{3}$, so $\frac{1}{64}=4^{-3}$, so $x=-3$. Now, is $-3$ a whole number? By the standard definition of whole numbers (non - negative integers), no. But maybe in the context of the problem, they consider integers. Wait, let's check the calculation again.

Wait, $4^{x}=\frac{1}{64}$. Let's solve for $x$:

$4^{x}=\frac{1}{64}\implies4^{x}=64^{-1}\implies4^{x}=(4^{3})^{-1}\implies4^{x}=4^{-3}\implies x=-3$.

Now, the problem says "determine if the value is a whole number or not". A whole number is a non - negative integer (0, 1, 2, 3,...). $-3$ is an integer but not a whole number. Wait, but maybe there is a miscalculation. Wait, no, the calculation is correct.

Wait, but let's check again:

$\log_{4}\frac{1}{64}$. Let's use the change - of - base formula: $\log_{4}\frac{1}{64}=\frac{\log\frac{1}{64}}{\log4}=\frac{\log1-\log64}{\log4}=\frac{0 - \log4^{3}}{\log4}=\frac{- 3\log4}{\log4}=-3$. So the value is $-3$, which is an integer but not a whole number (since whole numbers are non - negative).

# Answer:
$\log_{4}\frac{1}{64}$ is defined, and its value ($-3$) is not a whole number (because whole numbers are non - negative integers like 0, 1, 2, 3, ..., and $-3$ is negative).

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QUESTION IMAGE

Question

Explanation:

Problem 1a: Evaluate $\boldsymbol{\log_{4} 64}$

Step1: Recall the logarithm definition

Step2: Express 64 as a power of 4

Step3: Determine the value of the logarithm

Step1: Recall the logarithm definition

Step2: Express 32 as a power of 2

Step3: Determine the value of the logarithm

Step1: Recall the logarithm definition

Step2: Express 49 as a power of 7

Step3: Determine the value of the logarithm

Answer:

Problem 1b: Evaluate $\boldsymbol{\log_{2} 32}$