Exponential Equation Involving Base 5 and 25
<p>Given:</p>
<p>\(25^{x-1}=5^{2x-1}-100\)</p>
<p>Since \(25\) is a power of \(5\), i.e., \(25 = 5^2\), rewrite the equation:</p>
<p>\((5^2)^{x-1}=5^{2x-1}-100\)</p>
<p>Apply the exponent product rule \( (a^m)^n = a^{mn} \):</p>
<p>\(5^{2(x-1)}=5^{2x-1}-100\)</p>
<p>\(5^{2x-2}=5^{2x-1}-100\)</p>
<p>Now, set the exponents of base \(5\) equal to each other and solve the resulting equation:</p>
<p>\(2x-2=2x-1\)</p>
<p>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:</p>
<p>Let \(y=5^{x-1}\), this gives us the following equation: </p>
<p>\(y = 5y - 100\)</p>
<p>Solve for \(y\):</p>
<p>\(100 = 5y - y\)</p>
<p>\(100 = 4y\)</p>
<p>\(y = 25\)</p>
<p>Substituting back for \(y\):</p>
<p>\(5^{x-1} = 25\)</p>
<p>Since \(25=5^2\), we have:</p>
<p>\(5^{x-1} = 5^2\)</p>
<p>Thus, \(x-1 = 2\)</p>
<p>Finally, solve for \(x\):</p>
<p>\(x = 2 + 1\)</p>
<p>\(x = 3\)</p>
