Question

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Answer 1

We have
\[(\frac{1}{8})^{x-3}= 2\times 16 ^{2x+1}\]
8= 2^3 so
\[(\frac{1}{2^3})^{x-3}= 2\times 16 ^{2x+1}\]
so we get
\[(2^{-3})^{(x-3)}=2 \times 16^{2x+1}\]
we know 16=2^4 so we get now
\[2^{-3x+9}=2 \times (2^4)^{2x+1}\]
we get now
\[2^{-3x+9}=2 \times 2^{8x+4}\]
divide both sides by \(2^{8x+4}\)

we get
\[2^{-3x+9} \times 2^{-8x-4}=2\]
\[2^{-11x+5}=2^1\]
so
\[-11x+5=1\]
or
\[x=-\frac{4}{11}\]

Answer 2

Thank you so much @ash2326 !!