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Logarithmic functions, such as the one seen in the exercise, \( y = 13 + \log x \), play a vital role in various areas of mathematics and science. They are the inverse functions of exponential functions and are used to express the power to which a certain base number must be raised to obtain another number.

For instance, in our case, the base is 10, as it is the common logarithm, and can be expressed as \( \log x = \log_{10} x \). When we see the notation \( \log x \), it means \( 10^n = x \), where \( n \) is the logarithm of \( x \) to the base 10. Logarithms help in simplifying complex calculations, especially when dealing with large numbers or exponential growth and decay, which are prevalent in fields like finance and natural sciences.

Exponential functions are functions of the form \( f(x) = b^{x} \), where \( b \) is a positive real number, and \( b eq 1 \). They describe situations where a quantity grows or decays at a rate proportional to its current value, making them essential in modeling real-world phenomena such as population growth, radioactive decay, and interest calculations.

In this exercise, we transformed a logarithmic equation into its inverse, the exponential form. The inverse of \( y = \log x \) is \( x = b^y \), which becomes \( y = b^x \) once we switch \( x \) and \( y \). Hence, through algebraic manipulation, we derived the inverse function \( y = 10^{(x - 13)} \) from the original logarithmic function.

Algebraic manipulation involves rearranging equations and expressions using algebraic operations to simplify them or to express them in a different but equivalent form. Such manipulations are crucial when finding inverse functions, as demonstrated in the given exercise.

Starting with the expression \( y = 13 + \log x \), we applied several steps of algebraic manipulation. Initially, we interchanged \( x \) and \( y \), yielding \( x = 13 + \log y \). Next, we isolated the logarithmic part by subtracting 13 from both sides, giving us \( x - 13 = \log y \). Finally, converting this into exponential form resulted in the inverse function, \( y = 10^{(x - 13)} \) where we've applied our understanding of logarithmic and exponential relationships, alongside algebraic steps, to solve for \( y \).