Let's go through each question step-by-step:
Question 1
A) If \( \log x = 0.25 \), we want to find \( \log x 4 \).
Using the power rule of logarithms, we can rewrite \( \log x 4 \) as: \[ \log x 4 = \log x (4) = \log x (4^1) = \log x (2^2) = 2 \log x 2 \] Since we already have \( \log x = 0.25 \), we can find \( \log x 2 \) by using the change of base formula: \[ \log x 2 = \frac{\log 2}{\log x} = \frac{\log 2}{0.25} = 4 \log 2 \] Let's substitute this back: \[ \log x 4 = 2 \cdot \log x (2) = 2 \cdot (4 \log 2) = 8 \log 2 \] To evaluate this, we need the value of \( \log 2 \). If not specified, let’s assume \( \log 2 \approx 0.301 \): \[ \log x 4 = 8 \cdot 0.301 \approx 2.408 \]
Question 2
A) Which of the following is equivalent to \( \log 9 \)?
Let's analyze each option:
- Option #1: \( \log(3+3) = \log(6) \) (not equivalent)
- Option #2: \( 3 \log 3 \) (logarithmic property does not apply here)
- Option #3: \( 2 \log 3 \) (since \( 9 = 3^2 \), we have \( \log 9 = \log 3^2 = 2 \log 3 \) = equivalent)
So the answer is Option #3: \( 2 \log 3 \).
Question 3
A) If \( \log h = 35 \), we want to find \( \log h 3 \).
Using the power rule of logarithms again: \[ \log h 3 = \log h (3^1) = \frac{\log 3}{\log h} \] Thus, \[ \log h 3 = \frac{\log 3}{35} \] This can be expressed as: \[ \log h 3 = \frac{1}{35} \log 3 \] Where \( \log 3 \) is an approximate value, unless specified otherwise in terms of a base.
Question 4
A) Evaluate \( \log \sqrt{100} 3 \). First, we know that: \[ \sqrt{100} = 10 \] So, we execute the logarithm: \[ \log_{10} 3 = \log 3 \] Taking it further, \[ \log \sqrt{100} 3 = \frac{\log 3}{\log 10} \] Since \( \log 10 = 1 \), we find: \[ \log \sqrt{100} 3 = \log 3 \approx 0.477 \]
Question 5
A) Find the value of \( \log 0.00001 \).
We recall that: \[ 0.00001 = 10^{-5} \] Thus, \[ \log 0.00001 = \log(10^{-5}) = -5 \]
Summary of Answers:
- \( 2.408 \text{ (approx.)} \)
- Option #3: \( 2 \log 3 \)
- \( \frac{\log 3}{35} \)
- \( 0.477 \text{ (approx.)} \)
- \( -5 \)