https://file.aiquickdraw.com/mathbot/file/1729143570254b8zg7jdn.jpg

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?

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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.

Learn how to solve the equation 25 * (2^(log(x))) = x using logarithmic and exponential properties. This step-by-step solution covers simplifying exponentials, logarithmic identities, and algebraic manipulation, suitable for Grades 10-12.

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