Algebraically solve the equation, $\left(4^{5 x+3}\right)^{2}=4$, for $x$. All answers should be exact. If there is more than one solution, separate them by commas. If there is no solution, type DNE.
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Answer

$\boxed{x=-0.5}$

Steps

Step 1 ：$\left(4^{5x+3}\right)^{2}=4^{2(5x+3)}=4^{10x+6}$

Step 2 ：$4=2^2$, so we can rewrite the equation as $(2^2)^{10x+6}=2^2$

Step 3 ：This simplifies to $2^{20x+12}=2^2$

Step 4 ：Since the bases are equal, the exponents must also be equal. Therefore, we can set up the equation $20x+12=2$

Step 5 ：Solving for $x$ gives $20x=2-12$

Step 6 ：$20x=-10$

Step 7 ：$x=-10/20$

Step 8 ：$\boxed{x=-0.5}$