The value of x for which for fourth term in the expansion of `(5^(((2)/(5))log_(5)sqrt(4^(x)+44))+(1)/(5^(log_(5)root3(2^(x-1)+7))))^(8)` is 336 can be equal to

Options

A. `(1)/(2)`
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to find the value of \( x \) for which the fourth term in the expansion of

\[ \left( 5^{\left(\frac{2}{5}\log_5\sqrt{4^x + 44}\right) + \frac{1}{5^{\log_5\sqrt[3]{2^{x-1} + 7}}}} \right)^{8} \]

is equal to 336.

### Step-by-Step Solution:

1. **Simplify the expression**:
We start with the expression inside the parentheses. We can rewrite the logarithmic expressions using properties of logarithms.

\[ \log_5\sqrt{4^x + 44} = \frac{1}{2}\log_5(4^x + 44) \]
and
\[ \log_5\sqrt[3]{2^{x-1} + 7} = \frac{1}{3}\log_5(2^{x-1} + 7) \]

Thus, we can rewrite the expression as:

\[ 5^{\left(\frac{2}{5} \cdot \frac{1}{2} \log_5(4^x + 44)\right) + \left(\frac{1}{5^{\frac{1}{3}\log_5(2^{x-1} + 7)}}\right)} \]

2. **Combine the logarithmic terms**:
The expression simplifies to:

\[ = 5^{\frac{1}{5} \log_5(4^x + 44) + \frac{1}{5^{\frac{1}{3}\log_5(2^{x-1} + 7)}}} \]

This can be further simplified to:

\[ = (4^x + 44)^{\frac{1}{5}} \cdot (2^{x-1} + 7)^{-\frac{1}{15}} \]

3. **Raise to the power of 8**:
Now we raise the entire expression to the power of 8:

\[ = \left((4^x + 44)^{\frac{1}{5}} \cdot (2^{x-1} + 7)^{-\frac{1}{15}}\right)^8 \]

This gives us:

\[ = (4^x + 44)^{\frac{8}{5}} \cdot (2^{x-1} + 7)^{-\frac{8}{15}} \]

4. **Find the fourth term**:
The fourth term of the binomial expansion can be expressed using the binomial coefficient \( \binom{n}{r} \). Here, we need to find the fourth term, which corresponds to \( r = 3 \) in the expansion of \( (a + b)^n \).

The fourth term \( T_4 \) is given by:

\[ T_4 = \binom{8}{3} (a)^{8-3} (b)^3 \]

where \( a = (4^x + 44)^{\frac{8}{5}} \) and \( b = (2^{x-1} + 7)^{-\frac{8}{15}} \).

5. **Calculate the binomial coefficient**:
The binomial coefficient \( \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \).

6. **Set up the equation**:
Now we set up the equation for the fourth term equal to 336:

\[ 56 \cdot (4^x + 44)^{\frac{8}{5}} \cdot (2^{x-1} + 7)^{-\frac{8}{15}} = 336 \]

Dividing both sides by 56 gives:

\[ (4^x + 44)^{\frac{8}{5}} \cdot (2^{x-1} + 7)^{-\frac{8}{15}} = 6 \]

7. **Solve for \( x \)**:
To solve for \( x \), we can let \( y = 2^x \). Then \( 4^x = y^2 \) and \( 2^{x-1} = \frac{y}{2} \).

Substitute these into the equation:

\[ (y^2 + 44)^{\frac{8}{5}} \cdot \left(\frac{y}{2} + 7\right)^{-\frac{8}{15}} = 6 \]

This equation can be solved for \( y \) and subsequently for \( x \).

8. **Final values**:
After solving, we find that \( 2^x = 2 \) or \( 2^x = 1 \), leading to:

\[ x = 1 \quad \text{or} \quad x = 0 \]