Logarithms To The Rescue?

$\begin{aligned} 100+25^{x-1} & = 5^{2x-1} \\ 100 + \frac {25^x}{25} & = \frac {5^{2x}}5 & \small \color{#3D99F6} \text{Multiply both sides by }25 \\ 2500 + 25^x & = 5(25^x) \\ \implies 4(25^x) & = 2500 & \small \color{#3D99F6} \text{Divide both sides by }4 \\ 25^x & = 25^2 \\ \implies x & = \boxed 2 \end{aligned}$

$\large \displaystyle 100 + 25^{x-1} = 5^{2x-1}$

$\large \displaystyle \implies 100 + (5^2)^{x-1} = 5^{2x-1}$

$\large \displaystyle \implies 100 + 5^{2(x-1)} = 5^{2x-1}$

$\large \displaystyle \implies 100 + 5^{2x-2} = 5^{2x-1}$

$\large \displaystyle \implies 100 + 5^{2x-2} = 5(5^{2x-2})$

$\large \displaystyle \implies 5(5^{2x-2}) - (5^{2x-2})= 100$

$\large \displaystyle \implies 4(5^{2x-2}) = 100$

$\large \displaystyle \implies 5^{2x-2} = 25$

$\large \displaystyle \implies 5^{2x-2} = 5^2$

Since $5^x$ is an increasing function on $R$ , it is one-one on $R$ . Hence, $5^a = 5^b \implies a = b$ for real numbers $a$ and $b$ . Therefore,

$\large \displaystyle \implies 2x-2 = 2 \implies \boxed{x = 2}$

$\begin{aligned} 5^{2x - 1} - 25^{x - 1} & = 100 \\ \Rightarrow 5^{2x} \cdot 5^{- 1} - 25^x \cdot 25^{-1} & = 100 \\ \Rightarrow 25^x \cdot \dfrac 15 -25^x \cdot \dfrac 1{25} & = 100 \\ \Rightarrow 25^x\left(\dfrac 15 - \dfrac 1{25} \right) & = 100 \\ \Rightarrow 25^x\left(\dfrac4{25} \right) & = 100 \\ \Rightarrow 25^x & = \dfrac{100}4 \times 25 \\ \Rightarrow 25^x & = 25 \times 25 \\ \Rightarrow 25^x & = 25^2 \\ \implies x & = \boxed 2 \\ \end{aligned}$