IF ` 20 ^(3- 2x^2 = ( 40 sqrt(5) ) ^(3x^2-2)`, then x=

Options

A. `+- sqrt((13)/(12))`
B. `+- sqrt((12)/(13))`
C. `+- sqrt((4)/(5))`
D. `+- sqrt((5)/(4))`

Answer

To solve the equation \( 20^{3 - 2x^2} = (40 \sqrt{5})^{3x^2 - 2} \), we will follow these steps:

### Step 1: Rewrite the bases
We know that \( 40 \sqrt{5} \) can be expressed in terms of \( 20 \):
\[ 40 \sqrt{5} = 20 \cdot 2 \sqrt{5} = 20 \cdot \sqrt{20} \]
Thus, we can rewrite the equation as:
\[ 20^{3 - 2x^2} = (20 \cdot \sqrt{20})^{3x^2 - 2} \]

### Step 2: Simplify the right-hand side
Now we can express \( (20 \cdot \sqrt{20})^{3x^2 - 2} \):
\[ (20 \cdot \sqrt{20})^{3x^2 - 2} = 20^{3x^2 - 2} \cdot (\sqrt{20})^{3x^2 - 2} \]
Since \( \sqrt{20} = 20^{1/2} \), we have:
\[ (\sqrt{20})^{3x^2 - 2} = (20^{1/2})^{3x^2 - 2} = 20^{(3x^2 - 2)/2} \]
Thus, the right-hand side becomes:
\[ 20^{3x^2 - 2} \cdot 20^{(3x^2 - 2)/2} = 20^{3x^2 - 2 + (3x^2 - 2)/2} \]

### Step 3: Combine the exponents
To combine the exponents on the right-hand side:
\[ 3x^2 - 2 + \frac{3x^2 - 2}{2} = 3x^2 - 2 + \frac{3x^2}{2} - 1 = \frac{6x^2 - 4 + 3x^2 - 2}{2} = \frac{9x^2 - 6}{2} \]
So we can rewrite the equation as:
\[ 20^{3 - 2x^2} = 20^{\frac{9x^2 - 6}{2}} \]

### Step 4: Set the exponents equal to each other
Since the bases are the same, we can set the exponents equal to each other:
\[ 3 - 2x^2 = \frac{9x^2 - 6}{2} \]

### Step 5: Clear the fraction
To eliminate the fraction, multiply both sides by 2:
\[ 2(3 - 2x^2) = 9x^2 - 6 \]
This simplifies to:
\[ 6 - 4x^2 = 9x^2 - 6 \]

### Step 6: Rearrange the equation
Now, rearranging gives:
\[ 6 + 6 = 9x^2 + 4x^2 \]
\[ 12 = 13x^2 \]

### Step 7: Solve for \( x^2 \)
Dividing both sides by 13:
\[ x^2 = \frac{12}{13} \]

### Step 8: Solve for \( x \)
Taking the square root of both sides, we find:
\[ x = \pm \sqrt{\frac{12}{13}} = \pm \frac{2\sqrt{3}}{\sqrt{13}} \]

### Final Answer
Thus, the value of \( x \) is:
\[ x = \pm \frac{2\sqrt{3}}{\sqrt{13}} \]

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