Day-8 Each One Teach One (8) Find the value of $$ \log _{10} 0.0001 $$ (9) Sol: Let $$ \log _{10} 0.0001=x $$ 10 Remove log $$ \begin{array}{l} \text { 0. } 0001=10^{x} \ \Rightarrow \frac{1}{10,000}=10 \frac{1}{104} \ \Rightarrow \frac{1}{104}=10^{x} \ \Rightarrow 10^{-4}=10^{x} \ \Rightarrow 10^{x}=10^{-4} \end{array} $$ ( .. If Bases are equal, then Exponents are also equal.) $$ \therefore \log _{10} 0.0001=-4 $$ (1) Find the value $$ 3^{3}+\log { }^{3} $$. (2) Find the value $$ \sqrt{x} \log \frac{\sqrt{x}}{\sqrt{x}} $$. (3) Find the value $$ \log \sqrt{ } x $$. 1 (4) Show that $$ \log \frac{162}{343}+2 \log \frac{7}{9}-\log \frac{1}{7}=\log 2 $$.

Question

UpStudy AI · Accepted Answer

Sure, let's solve each of the given problems step by step.

---

### Problem (4)
**Show that:**
$$
\log \left( \frac{162}{343} \right) + 2 \log \left( \frac{7}{9} \right) - \log \left( \frac{1}{7} \right) = \log 2
$$

**Solution:**

We'll use logarithm properties to simplify the left-hand side (LHS) of the equation:

1. **Apply the Quotient Rule:**
   $$
   \log \left( \frac{a}{b} \right) = \log a - \log b
   $$
   
2. **Apply the Power Rule:**
   $$
   k \cdot \log a = \log a^k
   $$
   
3. **Apply the Product Rule as needed:**

Let's break down the LHS step by step.

**Step 1: Expand Each Logarithm Using the Quotient and Power Rules**
$$
\begin{align*}
\text{LHS} &= \log \left( \frac{162}{343} \right) + 2 \log \left( \frac{7}{9} \right) - \log \left( \frac{1}{7} \right) \
&= (\log 162 - \log 343) + 2 (\log 7 - \log 9) - (\log 1 - \log 7) \
\end{align*}
$$

**Step 2: Simplify Using Known Logarithm Values**
$$
\log 1 = 0 \quad \text{(since } 10^0 = 1\text{)}
$$
$$
\begin{align*}
\text{LHS} &= \log 162 - \log 343 + 2\log 7 - 2\log 9 - 0 + \log 7 \
&= \log 162 - \log 343 + 3\log 7 - 2\log 9 \
\end{align*}
$$

**Step 3: Express Numbers as Powers of Primes**
$$
162 = 2 \times 81 = 2 \times 3^4 \
343 = 7^3 \
9 = 3^2 \
$$
$$
\begin{align*}
\text{LHS} &= \log (2 \times 3^4) - \log (7^3) + 3\log 7 - 2\log (3^2) \
&= \log 2 + 4\log 3 - 3\log 7 + 3\log 7 - 4\log 3 \
\end{align*}
$$

**Step 4: Combine Like Terms**
$$
\begin{align*}
\text{LHS} &= (\log 2) + (4\log 3 - 4\log 3) + (-3\log 7 + 3\log 7) \
&= \log 2 + 0 + 0 \
&= \log 2 \
\end{align*}
$$

**Conclusion:**
$$
\log \left( \frac{162}{343} \right) + 2 \log \left( \frac{7}{9} \right) - \log \left( \frac{1}{7} \right) = \log 2
$$

---

### Additional Problems

#### Problem (2)
**Find the value of:**
$$
\sqrt{x} \cdot \log \left( \frac{\sqrt{x}}{\sqrt{x}} \right)
$$

**Solution:**

Simplify the expression inside the logarithm first:
$$
\frac{\sqrt{x}}{\sqrt{x}} = 1
$$
$$
\log(1) = 0 \quad \text{(since } 10^0 = 1\text{)}
$$
$$
\sqrt{x} \cdot 0 = 0
$$

**Conclusion:**
$$
\sqrt{x} \cdot \log \left( \frac{\sqrt{x}}{\sqrt{x}} \right) = 0
$$

---

#### Problem (3)
**Find the value of:**
$$
\log (\sqrt{x})
$$

**Solution:**

Use the Power Rule of logarithms:
$$
\log (\sqrt{x}) = \log (x^{1/2}) = \frac{1}{2} \log x
$$

**Conclusion:**
$$
\log (\sqrt{x}) = \frac{1}{2} \log x
$$

---

#### Problem (1)
**Find the value of:**
$$
3^{3} + \log^{3}
$$

**Solution:**

It appears there's a typographical error in the expression $$\log^{3}$$. Typically, the logarithm function requires an argument, such as $$\log a$$. Assuming the intended expression is:
$$
3^{3} + (\log a)^{3}
$$
Then:
$$
3^{3} = 27
$$
$$
\therefore 3^{3} + (\log a)^{3} = 27 + (\log a)^{3}
$$
If a specific value for $$a$$ is provided, you can substitute it in to find the numerical value. Otherwise, the expression remains in terms of $$\log a$$.

**Conclusion:**
$$
3^{3} + \log^{3} = 27 + (\log a)^{3} \quad \text{(assuming a missing argument in the logarithm)}
$$

*If the expression was meant to be different, please provide the complete and accurate form so it can be solved accordingly.*

---

Feel free to ask if you need further clarification or assistance with any other problems!
$$ \log \left( \frac{162}{343} \right) + 2 \log \left( \frac{7}{9} \right) - \log \left( \frac{1}{7} \right) = \log 2 $$ $$ 3^{3} + \log^{3} = 27 + (\log a)^{3} $$ $$ \sqrt{x} \cdot \log \left( \frac{\sqrt{x}}{\sqrt{x}} \right) = 0 $$ $$ \log (\sqrt{x}) = \frac{1}{2} \log x $$

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