Let's solve each problem step by step.
**Problem 1: Evaluate log base 23 of 2346 to 3 decimal places.**
Firstly, to compute the logarithm of a number with an unusual base, we can use the change of base formula. The change of base formula allows us to convert the logarithm to a base that is more common, like base 10 or base e (natural logarithm).
The formula is:
\[ \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \]
where
- \( \log_b(a) \) is the logarithm we want to find (in base b of a),
- \( c \) is a new base to which we're changing, and
- \( a \) and \( b \) are the original number and base.
In our case, \( a = 2346 \) and \( b = 23 \), and we can take \( c \) to be the base of natural logarithms (base e or \( \log \) in calculators).
Using the formula we have:
\[ \log_{23}(2346) = \frac{\log(2346)}{\log(23)} \]
The logarithms can be computed using a calculator or logarithm tables, but for this explanation, I'll assume that we are using a calculator that can find natural logarithms.
After finding the natural logarithms (to a sufficient number of decimal places to maintain accuracy), we perform the division and then round the result to 3 decimal places.
Let’s say we find the values (using a calculator):
\[ \log(2346) \approx 7.760 \]
\[ \log(23) \approx 3.135 \]
Now we divide the two:
\[ \log_{23}(2346) \approx \frac{7.760}{3.135} \]
After division, we might get something like:
\[ \log_{23}(2346) \approx 2.475 \]
And to 3 decimal places, it would be \( 2.475 \), not rounded since the thousandths place is already less than 5.
**Problem 2: Find the value of a piece of equipment after 5 years of 23% yearly depreciation.**
The formula for depreciation at a constant rate is given by:
\[ V = P(1 - r)^n \]
where:
- \( V \) is the future value of the equipment after \( n \) years,
- \( P \) is the initial value of the equipment,
- \( r \) is the annual depreciation rate (expressed as a decimal, so 23% becomes 0.23),
- \( n \) is the number of years.
For our problem:
- \( P = \$8,000 \)
- \( r = 23\% = 0.23 \)
- \( n = 5 \) years
Let’s calculate the value after 5 years:
\[ V = 8000 \times (1 - 0.23)^5 \]
\[ V = 8000 \times (0.77)^5 \]
We need to calculate \( (0.77)^5 \) and multiply by 8000.
\[ (0.77)^5 \approx 0.275 \] (Using a calculator)
Now we multiply by the initial value:
\[ V \approx 8000 \times 0.275 \]
\[ V \approx 2200 \]
So, the value of the equipment after 5 years would be approximately \$2,200, assuming continuous depreciation at a constant rate.
