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When dealing with logarithmic equations, it's often useful to convert them into exponential form. This transformation helps simplify the process of solving for the unknown variable. In the given problem, we start with the logarithmic equation: \( \text{log}(275 x^2) = 38 \). Remember, the common logarithm \( \text{log} \) has a base of 10. So, the equation can be rewritten using the exponential form: \( 10^{38} = 275 x^2 \). This step is crucial because it moves the unknown term out of the logarithmic function and into a more solvable form. If you're ever stuck with a logarithmic equation, consider converting it to exponential form to clarify the relationship between the numbers and the variables.

Once in exponential form, our task is to solve for the variable \( x \). We have \( 10^{38} = 275 x^2 \) from the previous step. Here, we need to isolate \( x \), but first, we focus on simplifying the equation. This involves dividing both sides by 275 to get \( x^2 \) by itself: \( x^2 = \frac{10^{38}}{275} \). Now we need to solve for \( x \), which requires taking the square root of both sides of the equation. This gives us \( x = \pm \sqrt{\frac{10^{38}}{275}} \). Remember, squaring a number can return both a positive and a negative result, hence the \( \pm \).

Isolating the variable is a critical step in solving equations. In this exercise, after converting our logarithmic equation to exponential form, we isolated \( x^2 \) by dividing both sides of the equation by 275: \( x^2 = \frac{10^{38}}{275} \). By isolating \( x^2 \), we simplified the expression and made it easier to solve for \( x \). The final step of taking the square root of both sides provided the solutions for \( x \). This method of isolating the variable by performing inverse operations (like division or taking square roots) is a key strategy in algebra to make solving equations more manageable. Always look to isolate the variable as early in the process as possible, simplifying each step methodically.