We start by trying to solve the equation 8^x = 27 for x. Since 8 is equal to 2^3 and 27 is equal to 3^3, we can rewrite that equation as (2^3)^x = 3^3. Simplifying that using the rules of exponents results in 2^(3x) = 3^3.
Next, we take log base 2 of both sides to solve for x, which simplifies our equation to 3x = log2(3^3). Solving for x, we find x = log2(3^3) / 3. This comes out to approximately x = 1.584962500721156.
Now that we know x, we can find a value for 4^(2x-3). However, first we need to rewrite our equation using the rules of exponents. We start by recognizing that 4 can be written as 2^2, so we rewrite our equation as (2^2)^(2x-3) = 2^(2*(2x-3)) = 2^(4x-6).
Now we substitute the value of x we calculated earlier into this equation. This gives us 2^(4*(log2(3^3)/3)-6) which equals 2^(4*log2(3) - 6).
Now, using the properties of logarithms, we can simplify the exponential part: 4*log2(3) - 6 is equal to 2*log2(9) - 6 which equals log2(81) - 6. This can be further simplified using the property log base a of b - log base a of c = log base a of b/c. So, we get log2(81/64).
Finally, that leads to our answer, 2^(log2(81/64)) which equals 81/64. So, 4^(2x-3) = 1.265625 when 8^x = 27.
To summarize, given the equation 8^x=27, the value of x is approximately 1.584962500721156, and substituting this value into the equation 4^(2x-3), we obtain the result 1.265625.