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Exponential equations are equations where the variable appears in an exponent. These equations often look like this: \(a^x = b\), where \(a\) and \(b\) are known numbers, and \(x\) is our unknown variable that needs solving. When we encounter exponential equations, we aim to isolate the variable in the exponent to solve for it.
It's not always easy to do this by inspection or simple arithmetic methods, especially when \(a\) and \(b\) are not powers of a common base. That's where logarithms come in handy.
Most exponential equations can be tackled by using logarithms because they allow us to "bring down" the variable from the exponent, converting the equation into one that is easier to manage. This is crucial when the equation involves numbers not easily resolvable by guessing or intuition.

Logarithms are the inverse operation to exponentiation. This makes them incredibly useful for solving exponential equations. The natural logarithm, written as \(ln\), is particularly common in mathematics, linking closely to the constant \(e\).
There are several key properties of logarithms that aid in solving equations with exponents:

• Product Rule: \(ln(ab) = ln(a) + ln(b)\)
• Quotient Rule: \(ln\left(\frac{a}{b}\right) = ln(a) - ln(b)\)
• Power Rule: \(ln(a^b) = b ln(a)\)

In our problem, \(3^x = 2\), applying these properties helps to transform the equation into \( x ln(3) = ln(2)\). We used the power rule to "move" the \(x\) in the exponent to a coefficient of the logarithm, making it much simpler to isolate the variable. This shift is what enables us to solve the previously complex exponential equation.

Solving equations without calculators may seem daunting, but it is possible by using mathematical properties and logical thinking.
The key step involves manipulating the equation using properties of logarithms and exponents to uncover the value of the variable.
For our equation \(3^x = 2\), we first applied the natural logarithm to both sides. Now the complex exponent is more approachable. With \(x\) out as a multiplier in \(x ln(3) = ln(2)\), division solves for \(x\) giving us \(x = \frac{ln(2)}{ln(3)}\).
While the final computation of \(x\) will yield a numerical answer that often requires a calculator for a decimal approximation, the crucial part is understanding the algebra that leads to this final step.
This practice strengthens problem-solving skills and deepens an understanding of mathematical properties, building confidence in handling similar equations in the future without immediately reaching for computational tools.