3) find $ \log_{7} 7^{11} $\
4) show that $ \frac{1}{2} \log_{3} 36 = \log_{2} + \log_{3} $\
5) evaluate $ \log_{10} + \log_{100} + \log_{1000} + \log_{10000} $\
6) patrick expanded $ 2 \log_{3} x^{3} w^{5} $ to $ 6 \log_{3} x + 5 \log_{3} w $. determine if he is correct. explain in complete sentences\
7) if $ \ln \frac{\sqrt{x}}{y^{5}} = a \ln x - b \ln y $, find the value of $ a $.\
8) if $ \log_{3} \frac{x^{2} w^{4}}{3 z^{5}} = a \log_{3} x - b \log_{3} z + c \log_{3} w $, find $ a + b + c $

Question

Accepted Answer

### Problem 3: Find $\log_{7} 7^{11}$
# Explanation:
## Step 1: Recall the logarithm power rule
The power rule of logarithms states that $\log_{a} a^{b} = b$ for any positive real number $a
eq 1$ and real number $b$.
## Step 2: Apply the power rule
In the expression $\log_{7} 7^{11}$, we have $a = 7$ and $b = 11$. Using the power rule, we get $\log_{7} 7^{11}=11$.
# Answer:
$11$

### Problem 4: Show that $\frac{1}{2}\log_{3} 36=\log_{3} 2+\log_{3} 3$
# Explanation:
## Step 1: Simplify the left - hand side (LHS)
First, we know that $36 = 4	imes9=2^{2}	imes3^{2}$. Using the power rule of logarithms $\log_{a}M^{n}=n\log_{a}M$, we can rewrite $\frac{1}{2}\log_{3}36$ as $\frac{1}{2}\log_{3}(2^{2}	imes3^{2})$.
Then, using the product rule of logarithms $\log_{a}(MN)=\log_{a}M+\log_{a}N$, we have $\frac{1}{2}\log_{3}(2^{2}	imes3^{2})=\frac{1}{2}(\log_{3}2^{2}+\log_{3}3^{2})$.
Applying the power rule again, $\frac{1}{2}(\log_{3}2^{2}+\log_{3}3^{2})=\frac{1}{2}(2\log_{3}2 + 2\log_{3}3)$.
Simplify the expression: $\frac{1}{2}(2\log_{3}2+2\log_{3}3)=\log_{3}2+\log_{3}3$.
## Step 2: Simplify the right - hand side (RHS)
The right - hand side is $\log_{3}2+\log_{3}3$. Using the product rule of logarithms $\log_{a}M+\log_{a}N=\log_{a}(MN)$, we get $\log_{3}(2	imes3)=\log_{3}6$.
Now, let's check the left - hand side again. $\frac{1}{2}\log_{3}36=\frac{1}{2}\log_{3}(6	imes6)=\frac{1}{2}(\log_{3}6+\log_{3}6)=\log_{3}6$. And the right - hand side $\log_{3}2+\log_{3}3=\log_{3}(2	imes3)=\log_{3}6$. So LHS = RHS.
# Answer:
Shown that $\frac{1}{2}\log_{3}36=\log_{3}2+\log_{3}3$ (since both sides equal $\log_{3}6$)

### Problem 5: Evaluate $\log_{10}10+\log_{10}100+\log_{10}1000+\log_{10}10000$
# Explanation:
## Step 1: Rewrite each term using the power of 10
We know that $10 = 10^{1}$, $100 = 10^{2}$, $1000 = 10^{3}$, $10000 = 10^{4}$.
So, $\log_{10}10=\log_{10}10^{1}$, $\log_{10}100=\log_{10}10^{2}$, $\log_{10}1000=\log_{10}10^{3}$, $\log_{10}10000=\log_{10}10^{4}$.
## Step 2: Apply the power rule of logarithms
Using the power rule $\log_{a}a^{b}=b$, we have:
$\log_{10}10^{1} = 1$, $\log_{10}10^{2}=2$, $\log_{10}10^{3}=3$, $\log_{10}10^{4}=4$.
## Step 3: Sum the values
Now, we sum these values: $1 + 2+3 + 4=\frac{(1 + 4)	imes4}{2}=10$ (using the formula for the sum of an arithmetic series $S_{n}=\frac{n(a_{1}+a_{n})}{2}$, where $n = 4$, $a_{1}=1$, $a_{n}=4$).
# Answer:
$10$

### Problem 6: Patrick expanded $2\log_{3}x^{3}w^{5}$ to $6\log_{3}x + 5\log_{3}w$. Determine if he is correct. Explain in complete sentences.
# Explanation:
## Step 1: Recall the product rule of logarithms
The product rule of logarithms states that $\log_{a}(MN)=\log_{a}M+\log_{a}N$ for $a>0,a
eq1,M>0,N>0$.
## Step 2: Apply the product rule to $\log_{3}(x^{3}w^{5})$
First, $\log_{3}(x^{3}w^{5})=\log_{3}x^{3}+\log_{3}w^{5}$ (by the product rule).
## Step 3: Apply the power rule of logarithms
The power rule of logarithms states that $\log_{a}M^{n}=n\log_{a}M$ for $a > 0,a
eq1,M>0,n\in R$.
Applying the power rule to $\log_{3}x^{3}$ and $\log_{3}w^{5}$, we get $\log_{3}x^{3}=3\log_{3}x$ and $\log_{3}w^{5}=5\log_{3}w$.
## Step 4: Multiply by 2
Now, we have $2\log_{3}(x^{3}w^{5})=2(\log_{3}x^{3}+\log_{3}w^{5})=2(3\log_{3}x + 5\log_{3}w)=6\log_{3}x+10\log_{3}w$.
Patrick's expansion is $6\log_{3}x + 5\log_{3}w$, which is incorrect because when we apply the power rule to $\log_{3}w^{5}$ and then multiply by 2, we should get $10\log_{3}w$ instead of $5\log_{3}w$.
# Answer:
Patrick is incorrect. When expanding $2\log_{3}(x^{3}w^{5})$, first using the product rule $\log_{3}(x^{3}w^{5})=\log_{3}x^{3}+\log_{3}w^{5}$, then using the power rule $\log_{3}x^{3}=3\log_{3}x$ and $\log_{3}w^{5}=5\log_{3}w$, and finally multiplying by 2 gives $2	imes3\log_{3}x+2	imes5\log_{3}w = 6\log_{3}x + 10\log_{3}w$, not $6\log_{3}x+5\log_{3}w$.

### Problem 7: If $\ln\frac{\sqrt{x}}{y^{5}}=A\ln x - B\ln y$, find the value of $A$.
# Explanation:
## Step 1: Simplify $\ln\frac{\sqrt{x}}{y^{5}}$ using the quotient rule
The quotient rule of logarithms states that $\ln\frac{M}{N}=\ln M-\ln N$ for $M > 0,N>0$.
So, $\ln\frac{\sqrt{x}}{y^{5}}=\ln\sqrt{x}-\ln y^{5}$.
## Step 2: Simplify $\ln\sqrt{x}$ using the power rule
We know that $\sqrt{x}=x^{\frac{1}{2}}$. Using the power rule of logarithms $\ln M^{n}=n\ln M$, we have $\ln x^{\frac{1}{2}}=\frac{1}{2}\ln x$.
## Step 3: Simplify $\ln y^{5}$ using the power rule
Using the power rule, $\ln y^{5}=5\ln y$.
## Step 4: Compare with $A\ln x - B\ln y$
Now, $\ln\frac{\sqrt{x}}{y^{5}}=\frac{1}{2}\ln x-5\ln y$. Comparing with $A\ln x - B\ln y$, we can see that $A=\frac{1}{2}$.
# Answer:
$\frac{1}{2}$

### Problem 8: If $\log_{3}\frac{x^{2}w^{4}}{3z^{5}}=A\log_{3}x - B\log_{3}z + C\log_{3}w$, find $A + B + C$.
# Explanation:
## Step 1: Simplify $\log_{3}\frac{x^{2}w^{4}}{3z^{5}}$ using the quotient rule
The quotient rule of logarithms states that $\log_{a}\frac{M}{N}=\log_{a}M-\log_{a}N$ for $a>0,a
eq1,M>0,N>0$.
So, $\log_{3}\frac{x^{2}w^{4}}{3z^{5}}=\log_{3}(x^{2}w^{4})-\log_{3}(3z^{5})$.
## Step 2: Apply the product rule to $\log_{3}(x^{2}w^{4})$ and $\log_{3}(3z^{5})$
The product rule of logarithms states that $\log_{a}(MN)=\log_{a}M+\log_{a}N$ for $a>0,a
eq1,M>0,N>0$.
Applying the product rule:
$\log_{3}(x^{2}w^{4})=\log_{3}x^{2}+\log_{3}w^{4}$
$\log_{3}(3z^{5})=\log_{3}3+\log_{3}z^{5}$
## Step 3: Apply the power rule to each term
The power rule of logarithms states that $\log_{a}M^{n}=n\log_{a}M$ for $a>0,a
eq1,M>0,n\in R$.
Applying the power rule:
$\log_{3}x^{2}=2\log_{3}x$, $\log_{3}w^{4}=4\log_{3}w$, $\log_{3}3 = 1$, $\log_{3}z^{5}=5\log_{3}z$
## Step 4: Substitute back and simplify
$\log_{3}\frac{x^{2}w^{4}}{3z^{5}}=\log_{3}x^{2}+\log_{3}w^{4}-(\log_{3}3+\log_{3}z^{5})$
$=2\log_{3}x + 4\log_{3}w-1 - 5\log_{3}z$
$=2\log_{3}x-5\log_{3}z + 4\log_{3}w$
Comparing with $A\log_{3}x - B\log_{3}z + C\log_{3}w$, we have $A = 2$, $B = 5$, $C = 4$.
## Step 5: Calculate $A + B + C$
Now, $A + B + C=2 + 5+4 = 11$.
# Answer:
$11$

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QUESTION IMAGE

Question

Explanation:

Problem 3: Find $\log_{7} 7^{11}$

Step 1: Recall the logarithm power rule

Step 2: Apply the power rule

Step 1: Simplify the left - hand side (LHS)

Step 2: Simplify the right - hand side (RHS)

Step 1: Rewrite each term using the power of 10

Step 2: Apply the power rule of logarithms

Step 3: Sum the values

Answer:

Problem 4: Show that $\frac{1}{2}\log_{3} 36=\log_{3} 2+\log_{3} 3$