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A logarithmic function is essentially the inverse of an exponential function. If you have an equation involving a logarithm, like \text{ \( x = \log_{125} 5 \)}, it means you are trying to figure out what exponent you must raise 125 to in order to get 5.

Its general form is \text{ \( y = \log_b a \)}, where \text{ \( b \)} is the base, \text{ \( a \)} is the number you want to find a logarithm for, and \text{ \( y \)} is the exponent. In other words, you are asking: To what power must the base \text{ \( b \)} be raised to produce \text{ \( a \)}.

Here are a few useful properties of logarithms:

• \text{ \( \log_b(b^x) = x \)}
• Exponentiation reverses a logarithm: if \text{ \( y = \log_b a, \)}, then \text{ \( b^y = a \)}
• Logarithms turn multiplication into addition: \text{ \( \log_b(x \times y) = \log_b(x) + \log_b(y) \)}

Exponentiation is the process of raising a number to a power. In the equation \text{ \( 125^x = 5 \)}, exponentiation helps to identify the relationship between the base and the exponent.

In general terms, if you have \text{ \( a^b = x \)}, you are raising \text{ \( a \)} to the power of \text{ \( b \)} to get \text{ \( x \)}. In the context of logarithms, this is useful because logarithms and exponents are inverse operations.

Some key properties of exponents include:

• \text{ \( a^m \times a^n = a^{m+n} \)}
• \text{ \( (a^m)^n = a^{mn} \)}
• \text{ \( a^0 = 1 \)} for any non-zero \text{ \( a \)}

These properties are crucial when solving logarithmic equations by transforming them into exponential form.

Solving logarithmic equations often requires converting them to exponential form. Let's explore how to solve the given equation \text{ \( x = \log_{125} 5 \)} step by step.

First, acknowledge that the logarithmic equation means \text{ \( 125^x = 5 \)}. Next, find a common base for both numbers involved. We notice that 125 and 5 can be written as powers of 5: \text{ \(125 = 5^3 \)} and \text{ \(5 = 5^1 \)}.

By substituting these expressions, we get \text{ \( (5^3)^x = 5 \)} or, simplifying further, \text{ \( 5^{3x} = 5^1 \)}. With equal bases, the exponents must be equal, so \text{ \( 3x = 1 \)}.

Finally, solve for \text{ \( x \)} by dividing both sides by 3 to get \text{ \( x = \frac{1}{3} \)}. This step-by-step method makes it easier to understand and solve logarithmic equations by transforming them into a manageable exponential form.