The value of x, for which the ninth term in the
expansion of `{sqrt(10)/((sqrt(x))^(5log _(10)x ))+ x.x^(1/(2log_(10)x))}^(10)`
is 450 is equal to

Options

A. 10
B. `10^(2)`
C. `sqrt(10)`
D. `10^(-2//5)`

Explanation

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

\[ \left( \frac{\sqrt{10}}{(\sqrt{x})^{5 \log_{10} x}} + x \cdot x^{\frac{1}{2 \log_{10} x}} \right)^{10} \]

is equal to 450.

### Step 1: Simplify the expression
Let \( \log_{10} x = \lambda \). Then, we can express \( x \) as:

\[ x = 10^\lambda \]

Now, we can rewrite the expression inside the parentheses:

\[ \frac{\sqrt{10}}{(\sqrt{x})^{5 \log_{10} x}} = \frac{\sqrt{10}}{(\sqrt{10^\lambda})^{5 \lambda}} = \frac{\sqrt{10}}{(10^{\lambda/2})^{5 \lambda}} = \frac{\sqrt{10}}{10^{\frac{5 \lambda^2}{2}}} = \sqrt{10} \cdot 10^{-\frac{5 \lambda^2}{2}} \]

And for the second term:

\[ x \cdot x^{\frac{1}{2 \log_{10} x}} = 10^\lambda \cdot (10^\lambda)^{\frac{1}{2 \lambda}} = 10^\lambda \cdot 10^{\frac{1}{2}} = 10^{\lambda + \frac{1}{2}} \]

Thus, the expression simplifies to:

\[ \left( \sqrt{10} \cdot 10^{-\frac{5 \lambda^2}{2}} + 10^{\lambda + \frac{1}{2}} \right)^{10} \]

### Step 2: Identify the ninth term
The general term in the binomial expansion of \( (a + b)^n \) is given by:

\[ T_k = \binom{n}{k} a^{n-k} b^k \]

For our case, \( n = 10 \), \( a = \sqrt{10} \cdot 10^{-\frac{5 \lambda^2}{2}} \), and \( b = 10^{\lambda + \frac{1}{2}} \). The ninth term corresponds to \( k = 8 \):

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

Calculating \( T_9 \):

\[ T_9 = \binom{10}{8} \cdot 10^{1} \cdot 10^{-5 \lambda^2} \cdot 10^{8\lambda + 4} = \binom{10}{8} \cdot 10^{1 - 5 \lambda^2 + 8\lambda + 4} \]

### Step 3: Set the equation
We know that \( T_9 = 450 \):

\[ \binom{10}{8} \cdot 10^{5 - 5 \lambda^2 + 8\lambda} = 450 \]

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

\[ 45 \cdot 10^{5 - 5 \lambda^2 + 8\lambda} = 450 \]

Dividing both sides by 45:

\[ 10^{5 - 5 \lambda^2 + 8\lambda} = 10 \]

### Step 4: Solve for \( \lambda \)
This implies:

\[ 5 - 5 \lambda^2 + 8\lambda = 1 \]

Rearranging gives:

\[ -5 \lambda^2 + 8\lambda + 4 = 0 \]

Dividing through by -1:

\[ 5 \lambda^2 - 8\lambda - 4 = 0 \]

### Step 5: Use the quadratic formula
Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ \lambda = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 5 \cdot (-4)}}{2 \cdot 5} = \frac{8 \pm \sqrt{64 + 80}}{10} = \frac{8 \pm \sqrt{144}}{10} = \frac{8 \pm 12}{10} \]

Calculating the roots:

1. \( \lambda = \frac{20}{10} = 2 \)
2. \( \lambda = \frac{-4}{10} = -\frac{2}{5} \)

### Step 6: Find \( x \)
From \( \lambda = \log_{10} x \):

1. If \( \lambda = 2 \), then \( x = 10^2 = 100 \).
2. If \( \lambda = -\frac{2}{5} \), then \( x = 10^{-\frac{2}{5}} = 10^{-0.4} \).

Thus, the values of \( x \) are \( 100 \) and \( 10^{-0.4} \).

### Final Answer
The values of \( x \) are \( 100 \) and \( 10^{-0.4} \).

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