The problem is to solve the exponential equation $5 \times 2^{x-1} = 2 \times 5^{2x}$ for $x$.

1. Problem Description

The problem is to solve the exponential equation $5 \times 2^{x-1} = 2 \times 5^{2x}$ for $x$.

2. Solution Steps

First, divide both sides of the equation by 10, or equivalently, by $5 \times 2$:

$\frac{5 \times 2^{x-1}}{5 \times 2} = \frac{2 \times 5^{2x}}{5 \times 2}$

This simplifies to

$\frac{2^{x-1}}{2} = \frac{5^{2x}}{5}$

Then, using the rule $\frac{a^m}{a^n} = a^{m-n}$, we have

$2^{x-1-1} = 5^{2x-1}$

$2^{x-2} = 5^{2x-1}$

Now, take the logarithm of both sides (using any base, but the natural logarithm is convenient):

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

Using the logarithm property $\ln(a^b) = b \ln(a)$, we get

$(x-2)\ln(2) = (2x-1)\ln(5)$

$x \ln(2) - 2 \ln(2) = 2x \ln(5) - \ln(5)$

Now, rearrange the terms to isolate $x$:

$2x \ln(5) - x \ln(2) = \ln(5) - 2\ln(2)$

$x (2 \ln(5) - \ln(2)) = \ln(5) - 2\ln(2)$

So, we have

$x = \frac{\ln(5) - 2\ln(2)}{2 \ln(5) - \ln(2)}$

Using the property $a \ln(b) = \ln(b^a)$, we can rewrite the equation as

$x = \frac{\ln(5) - \ln(2^2)}{\ln(5^2) - \ln(2)}$

$x = \frac{\ln(5) - \ln(4)}{\ln(25) - \ln(2)}$

Using the property $\ln(a) - \ln(b) = \ln(\frac{a}{b})$, we get

$x = \frac{\ln(\frac{5}{4})}{\ln(\frac{25}{2})}$

Therefore,

$x = \frac{\ln(1.25)}{\ln(12.5)}$

Using a calculator, we can approximate the values:

$\ln(1.25) \approx 0.22314$

$\ln(12.5) \approx 2.52573$

$x \approx \frac{0.22314}{2.52573} \approx 0.08835$

However, let us stick to the exact form:

$x = \frac{\ln(5) - 2\ln(2)}{2\ln(5) - \ln(2)}$

3. Final Answer

$x = \frac{\ln(5) - 2\ln(2)}{2\ln(5) - \ln(2)}$