If log 10 1995 = 3.3000, then what is the value of (0.001995) 1/8 ?
Understanding the Logarithm Problem
The question asks us to find the value of \( (0.001995)^{1/8} \) given that \( \log_{10} 1995 = 3.3000 \). This problem involves using the properties of logarithms and exponents to simplify the expression and find its numerical value.
Relating 0.001995 to 1995
First, let's express the number \( 0.001995 \) in a way that relates it to \( 1995 \). We can write \( 0.001995 \) as:
\( 0.001995 = 1995 \times 0.000001 \)
Since \( 0.000001 \) is \( 10^{-6} \), we have:
\( 0.001995 = 1995 \times 10^{-6} \)
Applying the Given Logarithm Information
We are given that \( \log_{10} 1995 = 3.3000 \). By the definition of logarithm, this means that \( 10^{3.3000} = 1995 \). This is a crucial piece of information that will help us substitute the value of \( 1995 \) in terms of base 10.
Simplifying the Expression (0.001995)1/8
Now, let's substitute the expression for \( 0.001995 \) and the value of \( 1995 \) into the required expression \( (0.001995)^{1/8} \):
\( (0.001995)^{1/8} = (1995 \times 10^{-6})^{1/8} \)
Using the exponent property \( (ab)^c = a^c b^c \), we can separate the terms:
\( (1995 \times 10^{-6})^{1/8} = 1995^{1/8} \times (10^{-6})^{1/8} \)
Now, substitute \( 1995 = 10^{3.3000} \) into the expression:
\( (10^{3.3000})^{1/8} \times (10^{-6})^{1/8} \)
Using the exponent property \( (a^b)^c = a^{bc} \), we multiply the exponents:
\( 10^{(3.3000 \times \frac{1}{8})} \times 10^{(-6 \times \frac{1}{8})} \)
\( 10^{\frac{3.3000}{8}} \times 10^{\frac{-6}{8}} \)
Simplify the fractions in the exponents:
- \( \frac{3.3000}{8} \)
- \( \frac{-6}{8} = \frac{-3}{4} \)
So the expression becomes:
\( 10^{\frac{3.3000}{8}} \times 10^{-\frac{3}{4}} \)
Performing the Calculations
Let's calculate the values of the exponents:
- \( \frac{3.3000}{8} = 0.4125 \)
- \( -\frac{3}{4} = -0.75 \)
Substitute these values back into the expression:
\( 10^{0.4125} \times 10^{-0.75} \)
Using the exponent property \( a^m \times a^n = a^{m+n} \), we add the exponents:
\( 10^{0.4125 + (-0.75)} \)
\( 10^{0.4125 - 0.75} \)
Now, calculate the final exponent:
\( 0.4125 - 0.75 = -0.3375 \)
So the value of the expression is:
\( 10^{-0.3375} \)
Final Result in the Desired Format
The options are given in the form \( \frac{1}{10^x} \). We know that \( a^{-b} = \frac{1}{a^b} \). Therefore, we can write \( 10^{-0.3375} \) as:
\( \frac{1}{10^{0.3375}} \)
This matches one of the given options.
Comparison with Options
Let's compare our result with the provided options:
- \( \frac{1}{{{{10}^{0.3475}}}}\)
- \( \frac{1}{{{{10}^{0.3375}}}}\)
- \( \frac{1}{{{{10}^{0.3275}}}}\)
- \( \frac{1}{{{{10}^{0.3735}}}}\)
Our calculated value is \( \frac{1}{10^{0.3375}} \), which matches option 2.
| Step | Calculation / Reasoning | Expression |
|---|---|---|
| 1 | Rewrite 0.001995 | \( 0.001995 = 1995 \times 10^{-6} \) |
| 2 | Apply the power \( \frac{1}{8} \) | \( (0.001995)^{1/8} = (1995 \times 10^{-6})^{1/8} \) |
| 3 | Separate terms using \( (ab)^c \) rule | \( 1995^{1/8} \times (10^{-6})^{1/8} \) |
| 4 | Use given \( \log_{10} 1995 = 3.3000 \implies 1995 = 10^{3.3000} \) | \( (10^{3.3000})^{1/8} \times (10^{-6})^{1/8} \) |
| 5 | Apply \( (a^b)^c \) rule | \( 10^{\frac{3.3000}{8}} \times 10^{\frac{-6}{8}} \) |
| 6 | Simplify exponents | \( 10^{0.4125} \times 10^{-0.75} \) |
| 7 | Combine using \( a^m \times a^n \) rule | \( 10^{0.4125 - 0.75} \) |
| 8 | Calculate final exponent | \( 10^{-0.3375} \) |
| 9 | Rewrite using \( a^{-b} \) rule | \( \frac{1}{10^{0.3375}} \) |
Revision Table: Key Concepts
| Concept | Description | Formula/Rule |
|---|---|---|
| Logarithm Definition | If \( \log_b x = y \), then \( b^y = x \). | \( \log_b x = y \iff b^y = x \) |
| Product Rule for Exponents | When multiplying powers with the same base, add the exponents. | \( a^m \times a^n = a^{m+n} \) |
| Power of a Power Rule | To raise a power to a power, multiply the exponents. | \( (a^m)^n = a^{mn} \) |
| Power of a Product Rule | To raise a product to a power, raise each factor to the power. | \( (ab)^m = a^m b^m \) |
| Negative Exponent Rule | A term with a negative exponent is the reciprocal of the term with a positive exponent. | \( a^{-m} = \frac{1}{a^m} \) |
Additional Information: Extending Logarithm and Exponent Rules
Understanding logarithm and exponent rules is fundamental for solving such problems. These rules allow us to manipulate expressions involving powers and roots efficiently. For example, the rule \( (a^b)^c = a^{bc} \) is particularly useful when dealing with roots, as a root can be expressed as a fractional exponent, like \( x^{1/n} = \sqrt[n]{x} \).
The relationship between logarithms and exponents, \( \log_b x = y \iff b^y = x \), is the basis for converting between logarithmic and exponential forms, which was essential in this problem to use the given information \( \log_{10} 1995 = 3.3000 \).
When dealing with decimal numbers like \( 0.001995 \), it's often helpful to express them in scientific notation or as a product of an integer and a power of 10 (e.g., \( 1995 \times 10^{-6} \)) to make calculations clearer and easier, especially when applying exponent rules.
This problem demonstrates how different mathematical concepts (logarithms, exponents, decimal manipulation) are often interconnected and how applying their properties allows us to simplify complex expressions to a desired form.
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