Solve the following exponential equations algebraically. Show in
detail your workings and do the checks in detail to get full marks.
2(4^2𝑥+1) = 128
To solve the exponential equation 2(4^(2𝑥+1)) = 128 algebraically, we need to first simplify the equation by evaluating the exponential term.
Step 1: Simplify the exponential term
4^(2𝑥+1) = (2^2)^(2𝑥+1) = 2^(2(2𝑥+1)) = 2^(4𝑥+2)
Step 2: Substitute the simplified exponential term back into the equation
2(2^(4𝑥+2)) = 128
Step 3: Expand the expression by distributing the 2
2 * 2^(4𝑥+2) = 128
2^(4𝑥+3) = 128
Step 4: Convert both sides of the equation to the same base 2
2^(4𝑥+3) = 2^7
Step 5: Set the exponents equal to each other
4𝑥+3 = 7
4𝑥 = 4
𝑥 = 1
Step 6: Check the solution
2(4^(2(1)+1)) = 2(4^3) = 2(64) = 128
Therefore, 𝑥 = 1 is the solution to the exponential equation 2(4^(2𝑥+1)) = 128.