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 \( \log \frac{162}{343}+2 \log \frac{7}{9}-\log \frac{1}{7}=\log 2 \).

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!