91Ó°ÊÓ

Understanding how to convert fractions to exponents is crucial when solving exponential equations. In our exercise, we are confronted with the equation involving a base of 2 and a fractional number on the opposite side, \(2^{5x+3} = \frac{1}{8}\).

To proceed with the equation, we recognize that \(\frac{1}{8}\) can be rewritten using the base of 2 because 8 is a power of 2, precisely \2^3\. When a number is in the denominator, we convert it to a negative exponent in the numerator, thus \(\frac{1}{2^3}\) becomes \2^{-3}\. This is based on the exponent rule that states \(a^{-n} = \frac{1}{a^n}\).

By applying this principle, the equation now has a uniform base on both sides, allowing us to focus solely on the exponents for the next steps.

Once we've transformed both sides of the equation to have the same base, the next step is to equate the exponents. This principle stems from one of the fundamental properties of exponents, which says that if \a^b = a^c\, then \b = c\, as long as \a\ is not zero and both \b\ and \c\ are real numbers.

In our exercise, we've reached the point where the equation \(2^{5x+3}=2^{-3}\) allows us to set the exponents equal to each other, because the bases are already equal. By doing this, we obtain a linear equation \(5x + 3 = -3\) which is much simpler to solve. Equating exponents effectively converts an exponential equation into a linear one.

After equating exponents, we've derived a linear equation, which is an equation of the first degree where each term is either a constant or the product of a constant and a single variable. Linear equations look like \(ax + b = 0\), where \a\ and \b\ are constants.

In this case, we have \(5x + 3 = -3\). To solve this linear equation, we follow a systematic approach: isolate the variable \x\ by performing inverse operations. First, we subtract 3 from both sides to get \(5x = -6\). Then, we divide each side by 5, yielding the solution \(x = -\frac{6}{5}\).

Linear equations are core to algebra and understanding how to manipulate them is key. When faced with more complex algebraic equations, converting them to linear form, if possible, can make finding solutions significantly easier.