Robert Bosch Logarithms Quiz 3

Answer: C) 6.3

Explanation:
If \log(x) = 0.8, then x = 10^{0.8} \approx 6.3095.
Since the options are not exact, the closest value is 6.3.

Answer: B) 2.5
Explanation:
We know that \log(a \cdot b) = \log(a) + \log(b).
Therefore, \log(12) = \log(3 \cdot 4) = \log(3) + \log(4) \approx 1.0986 + 1.3863 \approx 2.4849 \approx 2.5 (rounded to one decimal place).

Answer: A) -1.699
Explanation: We know that \log\left(\frac{m}{n}\right) = \log(m) - \log(n). Therefore, \log\left(\frac{m}{n}\right) = -2.3010 - (-0.6021) = -1.6989 \approx -1.699 (rounded to three decimal places).

Answer: A) 1.100

Explanation:
We know that \log(a \cdot b) = \log(a) + \log(b) and \log(a / b) = \log(a) - \log(b).
Therefore, \log(5) + \log(20) - \log(8) = \log(5 \cdot 20 / 8) = \log(12.5) \approx 1.0969 \approx 1.100 (rounded to three decimal places).

Answer: D) 5.0575
Explanation:
We know that \log(a^m \cdot b^n) = m \cdot \log(a) + n \cdot \log(b).
Therefore, \log(p^2 \cdot q^3) = 2 \cdot \log(p) + 3 \cdot \log(q) = 2 \cdot (-0.1549) + 3 \cdot 1.7891 \approx 5.0575 (rounded to three decimal places).

Answer: C) 1728000
Explanation: If \log(x) = 2.0794, then x = 10^{2.0794} \approx 120.
To find x^3, we simply cube the value of x, i.e., x^3 \approx 120^3 \approx 1728000.

Answer: 3
Explanation: in this case, \log(64) / \log(4) is equal to \log_4(64).
Now, you need to find the exponent to which 4 must be raised to get 64:
4^x = 64
By observing, we see that 4^3 = 64.
Therefore, \log(64) / \log(4) = \log_4(64) = 3

Answer: C) 0.002637

Explanation: If \log(a) = -1.3010, then a = 10^{-1.3010} \approx 0.05012.
Therefore, a^2 + a^3 = (0.05012)^2 + (0.05012)^3 \approx 0.002512 + 0.00012591 \approx 0.002637 \approx 0.345 (rounded to three decimal places).

Answer: A) 0.301
Explanation: We know that \log(b/c) = \log(b) - \log(c).
Therefore, \log(b/c) = 0.6021 - 0.3010 = 0.3011 \approx 0.301 (rounded to three decimal places).

Answer: A) 1.5
Explanation: We know that \log(a \cdot b) = \log(a) + \log(b) and \log(a / b) = \log(a) - \log(b). Applying these properties, we get:
\log(1000) - \log(125) + \log(4) = \log(1000 / 125) + \log(4)
Now, simplify the inside of the logarithm:
1000 / 125 = 8
So, \log(1000 / 125) + \log(4) = \log(8) + \log(4)

Next, use the property of addition of logarithms:
\log(a) + \log(b) = \log(a \cdot b)
\log(8) + \log(4) = \log(8 \cdot 4) = \log(32)
Now, the value of \log(1000) - \log(125) + \log(4) is \log(32).
Finally, we can evaluate the logarithm:
\log(32) \approx 1.50515