$126_n = 86_{10}$
For the first problem, we are given the base-\(n\) representation \(126_n = 86_{10}\). By the definition of a base-\(n\) number, the expression \(126_n\) can be expanded as:
\[1 \cdot n^2 + 2 \cdot n^1 + 6 \cdot n^0 = n^2 + 2n + 6\]
Setting this equal to the decimal value \(86\), we have the quadratic equation:
\[n^2 + 2n + 6 = 86\]
Subtracting 86 from both sides gives:
\[n^2 + 2n - 80 = 0\]
We can solve this quadratic equation by factoring. We look for two numbers that multiply to \(-80\) and add to \(2\). These numbers are \(10\) and \(-8\). Thus, the equation factors as:
\[(n + 10)(n - 8) = 0\]
This gives two possible solutions: \(n = -10\) or \(n = 8\). Since the base \(n\) must be a positive integer greater than any of its digits (in this case, \(n > 6\)), the only valid solution is \(n = 8\).
For the second problem, we are asked to convert the exponential equation \(5^3 = 125\) into its logarithmic form. The definition of a logarithm states that for a base \(b\), an exponent \(y\), and a value \(x\), the equation \(b^y = x\) is equivalent to \(\log_b(x) = y\).
In this case, we have \(b = 5\), \(y = 3\), and \(x = 125\). Applying the definition, the logarithmic form is:
\[\log_5(125) = 3\]
This expression confirms that 5 raised to the power of 3 equals 125.
Understanding base systems is essential in computer science, where data is often represented in binary (base-2), octal (base-8), or hexadecimal (base-16). For example, knowing that \(126_8 = 86_{10}\) allows programmers to convert memory addresses or data values between different formats for debugging. Similarly, logarithms are fundamental in fields like acoustics (measuring decibels), chemistry (measuring pH levels), and economics (calculating compound interest growth), where they help simplify complex exponential relationships into linear ones.