Solving the Logarithm Equation Step-by-Step

We are asked to find the solution for the given logarithm equation:

\({\rm{lo}}{{\rm{g}}_{10}}{[1 - {\{ 1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}}\} ^{ - 1{\rm{\;}}}}]^{ - \frac{1}{2}}} = 1\)

To solve this equation, we need to isolate \(x\) by peeling away the layers of operations, starting from the outermost one.

Step 1: Convert Logarithmic Form to Exponential Form

The equation is in the form \({\rm{lo}}{{\rm{g}}_{b}}(a) = c\), which can be rewritten in exponential form as \(b^c = a\).

Here, the base \(b = 10\), the exponent \(c = 1\), and the argument \(a = {[1 - {\{ 1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}}\} ^{ - 1{\rm{\;}}}}]^{ - \frac{1}{2}}}\).

So, we get:

\(10^1 = {[1 - {\{ 1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}}\} ^{ - 1{\rm{\;}}}}]^{ - \frac{1}{2}}}\)

\(10 = {[1 - {\{ 1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}}\} ^{ - 1{\rm{\;}}}}]^{ - \frac{1}{2}}}\)

Step 2: Eliminate the Outer Exponent \(-\frac{1}{2}\)

To get rid of the exponent \(-\frac{1}{2}\) on the right side, we can raise both sides of the equation to the power of \(-2\) (since \(-\frac{1}{2} \times -2 = 1\)).

\(10^{-2} = {\left( {[1 - {\{ 1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}}\} ^{ - 1{\rm{\;}}}}]^{ - \frac{1}{2}}} \right)^{-2}}\)

\(\frac{1}{10^2} = [1 - {\{ 1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}}\} ^{ - 1{\rm{\;}}}}]^{ - \frac{1}{2} \times -2}\)

\(\frac{1}{100} = [1 - {\{ 1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}}\} ^{ - 1{\rm{\;}}}}]^{1}\)

\(\frac{1}{100} = 1 - {\{ 1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}}\} ^{ - 1{\rm{\;}}}}\)

Alternatively, we could square both sides first and then take the reciprocal, as done in the scratchpad:

\(10^2 = {\left( {[1 - {\{ 1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}}\} ^{ - 1{\rm{\;}}}}]^{ - \frac{1}{2}}} \right)^{2}}\)

\(100 = {[1 - {\{ 1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}}\} ^{ - 1{\rm{\;}}}}]^{ - 1}}\)

Now, take the reciprocal of both sides to deal with the power \(-1\):

\(\frac{1}{100} = 1 - {\{ 1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}}\} ^{ - 1{\rm{\;}}}}\)

Both approaches lead to the same result.

Step 3: Isolate the Next Term with Exponent \(-1\)

Rearrange the equation to isolate the term \({ \{ 1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}}\} ^{ - 1{\rm{\;}}}} \):

\( {\{ 1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}}\} ^{ - 1{\rm{\;}}}} = 1 - \frac{1}{100}\)

\( {\{ 1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}}\} ^{ - 1{\rm{\;}}}} = \frac{100 - 1}{100}\)

\( {\{ 1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}}\} ^{ - 1{\rm{\;}}}} = \frac{99}{100}\)

Step 4: Eliminate the Exponent \(-1\)

Take the reciprocal of both sides to remove the exponent \(-1\) on the left side:

\(1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}} = \frac{100}{99}\)

Step 5: Isolate the Term with \({\rm{x}}^2\)

Rearrange the equation to isolate the term \({\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}}\):

\( - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}} = \frac{100}{99} - 1\)

\( - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}} = \frac{100 - 99}{99}\)

\( - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}} = \frac{1}{99}\)

Multiply both sides by \(-1\):

\({\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}} = -\frac{1}{99}\)

Step 6: Eliminate the Exponent \(-1\) on \((1 - x^2)\)

Take the reciprocal of both sides:

\(1 - {\rm{\;}}{{\rm{x}}^2} = -99\)

Step 7: Solve for \({\rm{x}}^2\)

Isolate the \({\rm{\;}}{{\rm{x}}^2}\) term:

\(-{\rm{\;}}{{\rm{x}}^2} = -99 - 1\)

\(-{\rm{\;}}{{\rm{x}}^2} = -100\)

Multiply by \(-1\):

\({{\rm{x}}^2} = 100\)

Step 8: Solve for \({\rm{x}}\)

Take the square root of both sides:

\({{\rm{x}}} = \pm \sqrt{100}\)

\({{\rm{x}}} = \pm 10\)

So, the possible solutions are \(x = 10\) and \(x = -10\).

Step 9: Verify Solutions and Check Options

We should verify if these solutions are valid in the original equation, particularly checking if any terms become zero or negative where they shouldn't (like in the denominator of a fraction raised to -1 or the argument of a logarithm).

The terms raised to the power -1 require their bases not to be zero:

• \(1 - {{\rm{x}}^2} \neq 0 \implies {{\rm{x}}^2} \neq 1 \implies x \neq \pm 1\). Both \(x=\pm 10\) satisfy this.
• \(1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}} \neq 0\). From Step 4, this expression equals \(\frac{100}{99}\) for both \(x = \pm 10\), which is not zero. Satisfied.

The term raised to the power -1/2 requires its base to be positive for real solutions:

• \(1 - {\{ 1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}}\} ^{ - 1{\rm{\;}}}} > 0\). From Step 3, this expression equals \(\frac{99}{100}\) for both \(x = \pm 10\), which is positive. Satisfied.

The argument of the logarithm must be positive:

• \( {[1 - {\{ 1 - {\left( {1 - {\rm{\;}}{{\rm{x}}^2}} \right)^{ - 1}}\} ^{ - 1{\rm{\;}}}}]^{ - \frac{1}{2}}} > 0\). From Step 1, this expression equals 10 for both \(x = \pm 10\), which is positive. Satisfied.

Both \(x=10\) and \(x=-10\) are mathematically valid solutions based on the algebraic steps and domain constraints of real numbers. Comparing these with the provided options, we find that \(x = 10\) is listed as an option.

The solutions derived are \(x = 10\) and \(x = -10\).

Revision Table: Key Concepts Reviewed

Additional Information: Simplifying Expressions

When solving equations with nested expressions and negative exponents like this one, it's crucial to work from the outside in or simplify innermost parts first, depending on the structure. In this case, converting the logarithm and dealing with the outer powers first helped reveal simpler forms step by step.

Remember that \((a - b)^{-1} = \frac{1}{a-b}\) and \(a - b^{-1} = a - \frac{1}{b}\), which are different. Paying close attention to the parentheses and the base of the negative exponent is key to avoid mistakes in algebraic manipulation.

Always check potential solutions in the original equation to ensure they do not lead to undefined terms (like division by zero or logarithm of a non-positive number).