If 6 3 - 4x 4 x + 5 = 8 (Given log 10 2 = 0.301 and log 10 3 = 0.477), then which one of the following is correct?
1 < x < 2
Solving the Logarithmic Equation: \(6^{3-4x} \cdot 4^{x+5} = 8\)
The problem asks us to find the value of \(x\) in the equation \(6^{3 - 4x} \cdot 4^{x + 5} = 8\) and then determine which given range contains this value of \(x\). We are also provided with the approximate values for \(\log_{10} 2\) and \(\log_{10} 3\). To solve for \(x\), we will use the properties of logarithms.
Applying Logarithms to the Equation
We start by taking the logarithm base 10 on both sides of the equation:
\(\log_{10}(6^{3 - 4x} \cdot 4^{x + 5}) = \log_{10} 8\)
Using the logarithm property \(\log(ab) = \log a + \log b\), we can separate the terms on the left side:
\(\log_{10}(6^{3 - 4x}) + \log_{10}(4^{x + 5}) = \log_{10} 8\)
Next, we use the property \(\log(a^b) = b \log a\) to bring the exponents down as coefficients:
\((3 - 4x) \log_{10} 6 + (x + 5) \log_{10} 4 = \log_{10} 8\)
Expressing Terms in Base Logarithms (log 2, log 3)
We are given values for \(\log_{10} 2\) and \(\log_{10} 3\). Let's express \(\log_{10} 6\), \(\log_{10} 4\), and \(\log_{10} 8\) in terms of \(\log_{10} 2\) and \(\log_{10} 3\):
- \(\log_{10} 6 = \log_{10} (2 \times 3) = \log_{10} 2 + \log_{10} 3\) (using \(\log(ab) = \log a + \log b\))
- \(\log_{10} 4 = \log_{10} (2^2) = 2 \log_{10} 2\) (using \(\log(a^b) = b \log a\))
- \(\log_{10} 8 = \log_{10} (2^3) = 3 \log_{10} 2\) (using \(\log(a^b) = b \log a\))
Substitute these back into the equation:
\((3 - 4x) (\log_{10} 2 + \log_{10} 3) + (x + 5) (2 \log_{10} 2) = 3 \log_{10} 2\)
Substituting Given Log Values
Now, substitute the given approximate values: \(\log_{10} 2 = 0.301\) and \(\log_{10} 3 = 0.477\).
- \(\log_{10} 2 + \log_{10} 3 = 0.301 + 0.477 = 0.778\)
- \(2 \log_{10} 2 = 2 \times 0.301 = 0.602\)
- \(3 \log_{10} 2 = 3 \times 0.301 = 0.903\)
Substitute these numerical values into the equation:
\((3 - 4x) (0.778) + (x + 5) (0.602) = 0.903\)
Solving for x
Now we have a linear equation in terms of \(x\). Expand and solve for \(x\):
\(3 \times 0.778 - 4x \times 0.778 + x \times 0.602 + 5 \times 0.602 = 0.903\)
\(2.334 - 3.112x + 0.602x + 3.010 = 0.903\)
Combine the terms with \(x\) and the constant terms:
\((-3.112 + 0.602)x + (2.334 + 3.010) = 0.903\)
\(-2.510x + 5.344 = 0.903\)
Isolate the term with \(x\):
\(-2.510x = 0.903 - 5.344\)
\(-2.510x = -4.441\)
Solve for \(x\):
\(x = \frac{-4.441}{-2.510}\)
\(x = \frac{4.441}{2.510}\)
Calculating the value:
\(x \approx 1.77\)
Comparing x with the Given Ranges
We found that \(x\) is approximately \(1.77\). Now let's look at the options provided:
| Option | Range | Does \(x \approx 1.77\) fall in this range? |
|---|---|---|
| 1 | \(0 < x < 1\) | No (1.77 is greater than 1) |
| 2 | \(1 < x < 2\) | Yes (1.77 is between 1 and 2) |
| 3 | \(2 < x < 3\) | No (1.77 is less than 2) |
| 4 | \(3 < x < 4\) | No (1.77 is less than 3) |
The calculated value of \(x \approx 1.77\) falls within the range \(1 < x < 2\).
Conclusion
Based on our calculation, the value of \(x\) is approximately 1.77, which lies in the range \(1 < x < 2\).
Revision Table: Key Concepts for Solving Log Equations
| Concept | Description | Application in this Problem |
|---|---|---|
| Logarithm Definition | If \(b^y = x\), then \(\log_b x = y\). | Used implicitly when solving for the exponent \(x\). |
| Change of Base | \(\log_b a = \frac{\log_c a}{\log_c b}\). Not directly used, but important for logs. | The problem is conveniently in base 10, matching the given values. |
| Product Rule | \(\log_b (xy) = \log_b x + \log_b y\). | Applied to \(\log_{10}(6^{3-4x} \cdot 4^{x+5})\). |
| Power Rule | \(\log_b (x^y) = y \log_b x\). | Applied to \((3-4x)\log_{10}6\), \((x+5)\log_{10}4\), and \(\log_{10}8\). |
| Log of Base | \(\log_b b = 1\). | Not directly used here as we used \(\log_{10} 2\) and \(\log_{10} 3\). |
| Log of 1 | \(\log_b 1 = 0\). | Not used in this problem. |
| Solving Linear Equations | Basic algebraic steps to isolate the variable. | Used after substituting log values to find \(x\). |
Additional Information on Logarithms and Exponential Equations
Exponential equations are equations where the variable appears in the exponent. Logarithms are the inverse operation to exponentiation, making them essential tools for solving such equations, especially when the bases are not easily made equal.
- Why use \(\log_{10}\)? We chose base 10 logarithms because the given values (\(\log_{10} 2\) and \(\log_{10} 3\)) are in base 10. Any base could theoretically be used, but using base 10 simplifies substituting the given data.
- Approximation: The values \(\log_{10} 2 \approx 0.301\) and \(\log_{10} 3 \approx 0.477\) are approximations. Using more precise values would result in a slightly more accurate value for \(x\), but for the purpose of determining the range among the given options, these approximations are sufficient.
- General Approach: For solving \(a^{f(x)} = b^{g(x)}\), a common approach is to take the logarithm of both sides (using any convenient base, often natural log or base 10) and then use the power rule to turn the exponential equation into a potentially linear or polynomial equation in \(x\).
- Checking the Solution: While not required by the question, one could substitute the approximate value of \(x \approx 1.77\) back into the original equation \(6^{3 - 4x} \cdot 4^{x + 5}\) to see if it approximately equals 8. This is a good way to check your work.
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