If \(2^{2x-1}=\frac{1}{8^{x-3}}\), find \(x=?\)

Given: \(2^{2x-1}=\frac{1}{8^{x-3}} \)

Step 1: Use the concept of prime factorization to write the number as the product of its power:

\( 8=2\times 2\times 2\times 2=2^3\)

\(\Rightarrow 2^{2x-1}=\frac{1}{(2^3)^{x-3}} \)

Step 2: Use the appropriate laws of exponents to simplify the expressions in R.H.S:

\(\Rightarrow 2^{2x-1}=\frac{1}{(2)^{3(x-3)}} \qquad [  (a^m)^n=a^{mn}] \)

\( \Rightarrow 2^{2x-1}=(2)^{-3(x-3)} \qquad [ \frac{1}{a^m}=a^{-m}] \newline \Rightarrow 2x-1=-3(x-3)  \qquad [ a^m=a^n \iff m=n] \)

Step 3: Solve the equation for } x:

\( \Rightarrow 2x-1=-3x+9 \newline \Rightarrow 5x=10 \newline \Rightarrow x=\frac{10}{5} \)

\(\textbf{Therefore, } \quad x =2 \).