Solve the exponential equation: 2(52x) - 7(5^x) = 15.

The given equation is $2(5^{2x}) - 7(5^x) = 15$.
Let $y = 5^x$. Then $5^{2x} = (5^x)^2 = y^2$.
Substitute $y$ into the equation:

$$2y^2 - 7y = 15$$

Subtract 15 from both sides:

$$2y^2 - 7y - 15 = 0$$

We can factor the quadratic equation. We need two numbers that multiply to $2 \times (-15) = -30$ and add up to $-7$. These numbers are $-10$ and 3.
Rewrite the middle term:

$$2y^2 - 10y + 3y - 15 = 0$$

Factor by grouping:

$$2y(y - 5) + 3(y - 5) = 0$$ $$(2y + 3)(y - 5) = 0$$

This gives two possible solutions for $y$:

$$2y + 3 = 0 \implies 2y = -3 \implies y = -\frac{3}{2}$$ $$y - 5 = 0 \implies y = 5$$

Recall that $y = 5^x$. Since $5^x$ must always be a positive value for real $x$, we discard the negative solution for $y$.
Therefore, $y = 5$ is the only valid solution.
Set $5^x$ equal to 5:

$$5^x = 5$$

Since $5 = 5^1$, we can write:

$$5^x = 5^1$$

Equating the exponents, we find:

$$x = 1$$

The calculated value $x=1$ matches option D.

The final answer is $\text{D. 1}$.