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Exponential expressions are mathematical expressions where a number, known as the base, is raised to the power of an exponent. A typical example is \((1.5)^t\), where 1.5 is the base and \(t\) is the exponent. In the context of the original exercise, these expressions are used to model exponential growth or decay, such as compound interest, population growth, or in this case, an unspecified growth factor.When dealing with such expressions, the key is to manage the base and the exponent carefully. By isolating the exponential expression, as in Step 1 of the solution, you can more easily apply mathematical operations like logarithms. This isolation involves rearranging the equation, so the exponential expression stands alone on one side, making subsequent operations simpler and clearer.

The power rule of logarithms is a useful tool that simplifies expressions where an exponent is applied inside a logarithm. This rule can be expressed as: \( \ln(a^b) = b \times \ln(a) \). This allows us to bring the exponent from the argument of the logarithm and multiply it outside.When applied to the problem solution, after taking the natural logarithm of both sides \(\ln((1.5)^t)\), the power rule helps to remove the exponent from the base by rewriting the equation as \(t \cdot \ln(1.5)\). As a result, this transformation turns an exponential equation into a more manageable linear one. Using this rule is a crucial step in solving equations that involve exponents and makes logarithms a powerful tool in algebra.

Logarithms are essential when you need to solve equations involving exponential expressions. They provide a method to "undo" exponential functions, simplifying the process of solving for an unknown exponent. This is especially true when we're working with natural logarithms (\(\ln\)), which are particularly suited to equations involving the constant \(e\).In our problem, after isolating the exponential expression and applying the natural logarithm, we use the power rule to rewrite the equation so that \(t\) stands out. The final step involves solving for \(t\) by isolating it on one side of the equation. Dividing both sides by \(\ln(1.5)\) yields: \[ t = \frac{\ln(4)}{\ln(1.5)} \] This process shows the usefulness of logarithms in algebra, enabling us to turn complex multiplicative relationships into simpler additive or linear forms, thus making equations more straightforward to solve.