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Logarithmic equations come in handy when you need to solve equations involving logarithmic expressions. These equations are tied to their inverse operations, exponents. So, when we have a logarithmic equation, such as \( \log_{b}(x) = y \), it reads as "Log base \( b \) of \( x \) is equal to \( y \)." This actually means that \( x \) is \( b \) raised to the power of \( y \).
To transform a logarithmic equation into a more solvable exponential form, you can apply this principle. This transformation can help unveil solutions that may not be apparent in the logarithmic expression alone. Writing \( \log_{64}(x) = \frac{2}{3} \) as an exponential equation gives \( 64^{\frac{2}{3}} = x \).
Using this approach allows you to shift perspectives from the log equation to one where you can directly solve for \( x \) through exponentiation.

Once a logarithmic equation is converted into its exponential form, your next step is to solve for \( x \). This involves evaluating the number that results from the exponential expression.
Consider the converted equation \( 64^{\frac{2}{3}} = x \). It may initially look complex because of the fractional exponent, but breaking it down into simpler steps helps. Fractional exponents signify roots and powers; specifically, the denominator can be viewed as the root while the numerator indicates the power. Thus, \( 64^{\frac{2}{3}} \) is equivalent to \( (\sqrt[3]{64})^2 \).
By understanding and implementing this method, it becomes more straightforward to isolate and solve for \( x \). Evaluating \( \sqrt[3]{64} \) gives \( 4 \), and squaring it results in \( 16 \). Thus, \( x = 16 \). This step-by-step method can simplify seemingly complicated calculations.

The exponentiation rule is a powerful tool in mathematics, particularly when dealing with fractional exponents. It allows the transformation and simplification of mathematical expressions.
Consider \( a^{\frac{m}{n}} \). This expression can be decomposed using the exponentiation rule into \( (\sqrt[n]{a})^m \) or equivalently, \( \sqrt[n]{a^m} \). Both expressions provide clarity and ease in computation. For instance, taking \( 64^{\frac{2}{3}} \), this can be translated into \( \sqrt[3]{64^2} \), or \( (\sqrt[3]{64})^2 \).
Applying this rule, you first determine the cube root of 64, which is 4, and then square the result to get 16. Working through such problems with the exponentiation rule offers a systematic and less daunting way to arrive at solutions, reinforcing the connection between exponents and logarithms.