Sure, let's solve the given equation step by step:
[tex]\[ 2^{(5x - 1)} = 4 \times 2^{(3x + 1)} \][/tex]
First, let's simplify the right-hand side of the equation. We know that [tex]\(4\)[/tex] can be expressed as [tex]\(2^2\)[/tex], making the equation:
[tex]\[ 2^{(5x - 1)} = 2^2 \times 2^{(3x + 1)} \][/tex]
Using the property of exponents, [tex]\(a^m \times a^n = a^{m+n}\)[/tex], we combine the exponents on the right-hand side:
[tex]\[ 2^{(5x - 1)} = 2^{2 + 3x + 1} \][/tex]
Next, simplify the exponent on the right-hand side:
[tex]\[ 2^{(5x - 1)} = 2^{(3x + 3)} \][/tex]
Since the bases are the same, we can set the exponents equal to each other:
[tex]\[ 5x - 1 = 3x + 3 \][/tex]
To isolate x, we first subtract [tex]\(3x\)[/tex] from both sides of the equation:
[tex]\[ 2x - 1 = 3 \][/tex]
Next, add 1 to both sides:
[tex]\[ 2x = 4 \][/tex]
Then, divide both sides by 2:
[tex]\[ x = 2 \][/tex]
Now, there is another solution corresponding to the equation's periodic nature in complex domains. However, for the consistency of this level of algebra, we focus on the fundamental solution we calculated, as [tex]\( x = 2 \)[/tex].
Hence, the solution to the equation [tex]\( 2^{(5x - 1)} = 4 \times 2^{(3x + 1)} \)[/tex] is:
[tex]\[ \boxed{x = 2} \][/tex]
