If the sixth term in the binomial expansion of \(\left(x^{-\frac{8}{3}}+x^2 \log _{10} x\right)^8\) is 5600, then what is the value of x ?
10
Understanding the Binomial Expansion Problem
This problem asks us to find the value of \(x\) given information about a specific term in a binomial expansion. The binomial expression is \(\left(x^{-\frac{8}{3}}+x^2 \log _{10} x\right)^8\), and we are told that the sixth term in its expansion is equal to 5600.
To solve this, we need to use the general formula for a term in a binomial expansion and then set up an equation using the given value of the sixth term.
General Term in Binomial Expansion
The binomial theorem states that the expansion of \((a+b)^n\) is given by:
\[(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r\]
The \((r+1)^{\text{th}}\) term in this expansion is given by the formula:
\[T_{r+1} = \binom{n}{r} a^{n-r} b^r\]
In our problem, the binomial is \(\left(x^{-\frac{8}{3}}+x^2 \log _{10} x\right)^8\). Comparing this to \((a+b)^n\):
- \(a = x^{-\frac{8}{3}}\)
- \(b = x^2 \log _{10} x\)
- \(n = 8\)
Finding the Sixth Term (\(T_6\))
We are interested in the sixth term, which is \(T_6\). Using the formula \(T_{r+1}\), we set \(r+1 = 6\), which means \(r = 5\).
Now, substitute the values of \(n\), \(r\), \(a\), and \(b\) into the general term formula \(T_{r+1} = \binom{n}{r} a^{n-r} b^r\):
\[T_6 = T_{5+1} = \binom{8}{5} \left(x^{-\frac{8}{3}}\right)^{8-5} \left(x^2 \log _{10} x\right)^5\]
Let's simplify the terms:
- Calculate the binomial coefficient \(\binom{8}{5}\):
\[\binom{8}{5} = \binom{8}{8-5} = \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 1 \times 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 8 \times 7 = 56\]
- Simplify the term \(\left(x^{-\frac{8}{3}}\right)^{8-5}\):
\[\left(x^{-\frac{8}{3}}\right)^{3} = x^{(-\frac{8}{3}) \times 3} = x^{-8}\]
- Simplify the term \(\left(x^2 \log _{10} x\right)^5\):
\[\left(x^2 \log _{10} x\right)^5 = (x^2)^5 (\log _{10} x)^5 = x^{2 \times 5} (\log _{10} x)^5 = x^{10} (\log _{10} x)^5\]
Now, combine these parts to get the full expression for the sixth term:
\[T_6 = 56 \cdot x^{-8} \cdot x^{10} (\log _{10} x)^5\]
Combine the \(x\) terms using the rule \(x^m \cdot x^n = x^{m+n}\):
\[T_6 = 56 \cdot x^{-8+10} (\log _{10} x)^5\]
\[T_6 = 56 \cdot x^2 (\log _{10} x)^5\]
Setting Up the Equation to Solve for x
We are given that the sixth term is 5600. So, we can set our expression for \(T_6\) equal to 5600:
\[56 x^2 (\log _{10} x)^5 = 5600\]
To simplify, divide both sides of the equation by 56:
\[\frac{56 x^2 (\log _{10} x)^5}{56} = \frac{5600}{56}\]
\[x^2 (\log _{10} x)^5 = 100\]
Solving for x by Checking Options
We have the equation \(x^2 (\log _{10} x)^5 = 100\). We can test the given options for the value of \(x\) to see which one satisfies this equation.
- Option 1: \(x=6\)
Substitute \(x=6\) into the equation: \(6^2 (\log _{10} 6)^5 = 36 (\log _{10} 6)^5\). Since \(\log_{10} 6\) is approximately 0.778, \(36 \times (0.778)^5\) will be much less than 100.
- Option 2: \(x=8\)
Substitute \(x=8\) into the equation: \(8^2 (\log _{10} 8)^5 = 64 (\log _{10} 8)^5\). Since \(\log_{10} 8\) is approximately 0.903, \(64 \times (0.903)^5\) will be less than 100.
- Option 3: \(x=9\)
Substitute \(x=9\) into the equation: \(9^2 (\log _{10} 9)^5 = 81 (\log _{10} 9)^5\). Since \(\log_{10} 9\) is approximately 0.954, \(81 \times (0.954)^5\) will be less than 100.
- Option 4: \(x=10\)
Substitute \(x=10\) into the equation: \(10^2 (\log _{10} 10)^5\).
We know that \(\log_{10} 10 = 1\).
So, the expression becomes \(100 \times (1)^5 = 100 \times 1 = 100\).
The value \(x=10\) satisfies the equation \(x^2 (\log _{10} x)^5 = 100\).
Conclusion
Based on the calculation, the value of \(x\) that makes the sixth term of the binomial expansion \(\left(x^{-\frac{8}{3}}+x^2 \log _{10} x\right)^8\) equal to 5600 is 10.
| Term Number | r Value | General Term Formula | Our Expansion Terms |
|---|---|---|---|
| \(T_{r+1}\) | \(r\) | \(\binom{n}{r} a^{n-r} b^r\) | \(\binom{8}{r} (x^{-8/3})^{8-r} (x^2 \log_{10} x)^r\) |
| \(T_6\) | 5 | \(\binom{8}{5} a^{8-5} b^5\) | \(\binom{8}{5} (x^{-8/3})^3 (x^2 \log_{10} x)^5\) |
Revision Table: Key Concepts
| Concept | Description | Formula/Example |
|---|---|---|
| Binomial Theorem | Expands powers of a binomial \((a+b)^n\). | \((a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r\) |
| General Term | The \((r+1)^{\text{th}}\) term in the expansion. | \(T_{r+1} = \binom{n}{r} a^{n-r} b^r\) |
| Binomial Coefficient | \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\). Represents the number of ways to choose \(r\) items from a set of \(n\). | \(\binom{8}{5} = 56\) |
| Logarithm \(\log_{10} x\) | The power to which 10 must be raised to get \(x\). | \(\log_{10} 10 = 1\) (since \(10^1=10\)) |
| Exponent Rules | Rules for simplifying expressions with exponents. | \((x^m)^n = x^{mn}\), \(x^m x^n = x^{m+n}\) |
Additional Information: Related Concepts
Logarithms Base 10
A logarithm with base 10 is often written as \(\log_{10} x\) or sometimes just \(\log x\) (especially in calculators). It answers the question: "10 to what power equals \(x\)?"
- \(\log_{10} 1 = 0\) because \(10^0 = 1\)
- \(\log_{10} 10 = 1\) because \(10^1 = 10\)
- \(\log_{10} 100 = 2\) because \(10^2 = 100\)
In this problem, the property \(\log_{10} 10 = 1\) was crucial for checking the option \(x=10\).
Calculating Binomial Coefficients
The binomial coefficient \(\binom{n}{r}\) is read as "n choose r". It can be calculated using factorials:
\[\binom{n}{r} = \frac{n!}{r!(n-r)!}\]
Remember that \(n!\) (n factorial) is the product of all positive integers up to \(n\). For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Also, \(0! = 1\).
An important property is \(\binom{n}{r} = \binom{n}{n-r}\). This is why \(\binom{8}{5} = \binom{8}{3}\), which simplifies the calculation.
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