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Logarithmic functions are the inverse of exponential functions. If you know how to work with exponents, you can understand logarithms, because they 'undo' each other.
A logarithmic function is written as \(\text{log}_b(y) = x\), which means \((b^x = y)\). In simpler words, if you take \(\text{log}_2(64)\), it asks the question: 'To what power must 2 be raised to get 64?'
To understand this better, always remember that logarithms reverse exponentiation. It's like asking for the exponent given the base (2 in our example) and the result (64).
Logarithmic functions are very useful in solving equations where the unknown is in the exponent, such as in growth, decay, and interest calculations.

An exponent refers to the number of times a number (the base) is multiplied by itself. For example, in \((2^6)\), 2 is the base and 6 is the exponent. This means 2 is multiplied by itself 6 times (2 x 2 x 2 x 2 x 2 x 2) to get 64.
Exponents are fundamental in understanding logarithms, as logarithms are exponents found through the 'inverse' process.
Here are some key properties of exponents:

• \((a^m \times a^n = a^{m+n})\)
• \((a^m / a^n = a^{m-n})\)
• \((a^0 = 1)\), where a is not zero
• \((a^{-n} = 1/a^n)\)

Understanding these properties can make it easier to grasp how logarithms work, especially when converting between exponential and logarithmic forms.

Base 2 logarithms specifically use 2 as the base. In notation, it’s written as \(\text{log}_2(y)\). This tells us the power to which 2 must be raised to get y.
For example, \(\text{log}_2(64) = 6\) because \((2^6 = 64)\). This makes base 2 logarithms very useful in computer science, where binary (base 2) is extensively used.
Solving base 2 logarithms involves two main steps:

• Express the number (like 64) as a power of 2 (\(( 64 = 2^6 )\)).
• Find the exponent (like 6 in our example).

With these steps, you can easily find the value of any base 2 logarithm. For instance, if you encounter any number expressed as \((2^n)\), determining its base 2 logarithm should be straightforward.