Giải:
Có: \(\dfrac{2^{2x-3}}{4^{10}}=8^3.16^5\)
\(\Leftrightarrow2^{2x-3}=8^3.16^5.4^{10}\)
\(\Leftrightarrow2^{2x-3}=\left(2^3\right)^3.\left(2^4\right)^5.\left(2^2\right)^{10}\)
\(\Leftrightarrow2^{2x-3}=2^9.2^{20}.2^{20}\)
\(\Leftrightarrow2^{2x-3}=2^{9+20+20}\)
\(\Leftrightarrow2^{2x-3}=2^{49}\)
Vì \(2=2\)
Nên \(2x-3=49\)
\(\Leftrightarrow2x=49+3\)
\(\Leftrightarrow2x=52\)
\(\Leftrightarrow x=\dfrac{52}{2}\)
\(\Leftrightarrow x=26\)
Vậy \(x=26\)
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