The equation shown in the image is:

$25 \left( 2^{\log{x}} \right) = x$

Let’s solve it step by step.

Step 1: Simplify $2^{\log{x}}$

Using the property of logarithms: $a^{\log_a{x}} = x$ we can simplify $2^{\log{x}}$ to $x^{\log{2}}$, which is approximately $x^{0.3010}$ (since $\log{2} \approx 0.3010$).

Thus, the equation becomes: $25 \cdot x^{\log{2}} = x$

Step 2: Solve for $x$

We now have: $25 \cdot x^{0.3010} = x$ Divide both sides by $x^{0.3010}$: $25 = x^{1 - 0.3010} = x^{0.699}$ So we have: $25 = x^{0.699}$ Take the logarithm of both sides: $\log{25} = 0.699 \cdot \log{x}$ Since $\log{25} \approx 1.3979$, we get: $1.3979 = 0.699 \cdot \log{x}$ Solve for $\log{x}$: $\log{x} = \frac{1.3979}{0.699} \approx 2$ Thus, $x = 10^2 = 100$.

Solution:

The solution to the equation is $x = 100$.

Would you like further details or have any questions?

Here are five related questions to explore further:

1. How do logarithmic properties simplify exponent expressions like $2^{\log{x}}$?
2. Can the change-of-base formula for logarithms be used to solve similar equations?
3. What happens if the constant 25 is replaced by another number?
4. How do we solve for $x$ in different bases (e.g., natural logarithms)?
5. How can numerical methods help in solving logarithmic equations?

Tip: Always remember that logarithmic and exponential functions are inverses of each other. This relationship is crucial in simplifying and solving such equations.