Question - Exponential Equation Involving Base 5 and 25

Given:

\(25^{x-1}=5^{2x-1}-100\)

Since \(25\) is a power of \(5\), i.e., \(25 = 5^2\), rewrite the equation:

\((5^2)^{x-1}=5^{2x-1}-100\)

Apply the exponent product rule \( (a^m)^n = a^{mn} \):

\(5^{2(x-1)}=5^{2x-1}-100\)

\(5^{2x-2}=5^{2x-1}-100\)

Now, set the exponents of base \(5\) equal to each other and solve the resulting equation:

\(2x-2=2x-1\)

The above equation has no solution for \(x\) since \(2x-2\) cannot equal \(2x-1\). However, since it's a proof that without subtracting the \(100\), the equation would never hold true for any value of \(x\), we may have made an error. Instead, let's look for a different approach to this problem:

Let \(y=5^{x-1}\), this gives us the following equation:

\(y = 5y - 100\)

Solve for \(y\):

\(100 = 5y - y\)

\(100 = 4y\)

\(y = 25\)

Substituting back for \(y\):

\(5^{x-1} = 25\)

Since \(25=5^2\), we have:

\(5^{x-1} = 5^2\)

Thus, \(x-1 = 2\)

Finally, solve for \(x\):

\(x = 2 + 1\)

\(x = 3\)