If the ninth term in the expansion of `[3^(log_(3)sqrt(25^(x-1)+7))+3^(-1//8log_(3)(5^(x-1)+1))]^(10)` is equal to 180 and `x gt 1` then x is equal to

Options

A. `log_(e )15`
B. `log_(5)15`
C. `log_(10)15`
D. none

Answer

To solve the problem step by step, we need to find the value of \( x \) such that the ninth term in the expansion of

\[ \left[ 3^{\log_3 \sqrt{25^{x-1} + 7}} + 3^{-\frac{1}{8} \log_3 (5^{x-1} + 1)} \right]^{10} \]

is equal to 180, given that \( x > 1 \).

### Step 1: Simplify the terms inside the brackets

The first term can be simplified using the property of logarithms:

\[ 3^{\log_3 \sqrt{25^{x-1} + 7}} = \sqrt{25^{x-1} + 7} \]

The second term can also be simplified:

\[ 3^{-\frac{1}{8} \log_3 (5^{x-1} + 1)} = (5^{x-1} + 1)^{-\frac{1}{8}} \]

Thus, we can rewrite the expression as:

\[ \left[ \sqrt{25^{x-1} + 7} + (5^{x-1} + 1)^{-\frac{1}{8}} \right]^{10} \]

### Step 2: Identify the ninth term in the expansion

Using the binomial theorem, the \( r \)-th term in the expansion of \( (a + b)^n \) is given by:

\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]

For our case, we need the ninth term, which corresponds to \( r = 8 \):

\[ T_9 = \binom{10}{8} \left( \sqrt{25^{x-1} + 7} \right)^2 \left( (5^{x-1} + 1)^{-\frac{1}{8}} \right)^8 \]

### Step 3: Substitute the values

Calculating \( \binom{10}{8} = \binom{10}{2} = 45 \):

\[ T_9 = 45 \left( 25^{x-1} + 7 \right) \left( (5^{x-1} + 1)^{-\frac{1}{8}} \right)^8 \]

This simplifies to:

\[ T_9 = 45 \left( 25^{x-1} + 7 \right) \cdot (5^{x-1} + 1)^{-1} \]

### Step 4: Set the equation equal to 180

Now we set the expression for \( T_9 \) equal to 180:

\[ 45 \cdot \frac{25^{x-1} + 7}{5^{x-1} + 1} = 180 \]

### Step 5: Simplify the equation

Dividing both sides by 45:

\[ \frac{25^{x-1} + 7}{5^{x-1} + 1} = 4 \]

Cross-multiplying gives:

\[ 25^{x-1} + 7 = 4(5^{x-1} + 1) \]

Expanding the right side:

\[ 25^{x-1} + 7 = 4 \cdot 5^{x-1} + 4 \]

### Step 6: Rearranging the equation

Rearranging gives:

\[ 25^{x-1} - 4 \cdot 5^{x-1} + 3 = 0 \]

### Step 7: Substitute \( y = 5^{x-1} \)

Let \( y = 5^{x-1} \). Then \( 25^{x-1} = y^2 \):

\[ y^2 - 4y + 3 = 0 \]

### Step 8: Factor the quadratic equation

Factoring gives:

\[ (y - 3)(y - 1) = 0 \]

Thus, \( y = 3 \) or \( y = 1 \).

### Step 9: Solve for \( x \)

1. If \( y = 1 \):

\[ 5^{x-1} = 1 \implies x - 1 = 0 \implies x = 1 \quad (\text{not valid since } x > 1) \]

2. If \( y = 3 \):

\[ 5^{x-1} = 3 \implies x - 1 = \log_5 3 \implies x = \log_5 3 + 1 \]

### Final Step: Conclusion

Thus, the value of \( x \) is:

\[ x = 1 + \log_5 3 \]