To solve the equation ((2^x)-7)^2 = 1, we can follow these steps:
Step 1: Take the square root of both sides of the equation to remove the exponent.
√(((2^x)-7)^2) = √1.
Simplifying the square root of 1 on the right side gives:
2^x - 7 = ±1.
Step 2: Solve for two cases, one with the positive square root and another with the negative square root.
Case 1: 2^x - 7 = 1.
Adding 7 to both sides:
2^x = 8.
Step 3: Rewrite 8 as 2^3 since 2^3 = 8.
2^x = 2^3.
Step 4: Since the bases are equal, the exponents must be equal.
x = 3.
Case 2: 2^x - 7 = -1.
Adding 7 to both sides:
2^x = 6.
Step 5: We can see that there is no integer value for x that satisfies 2^x = 6. So, we need to find an approximate value.
Step 6: Take the logarithm of both sides of the equation with base 2 to solve for x.
log2(2^x) = log2(6).
Using the logarithmic property of exponents:
x * log2(2) = log2(6).
Since log2(2) equals 1:
x = log2(6).
Using a calculator, we can find that log2(6) ≈ 2.58496.
Step 7: Round the solution to three significant figures.
Therefore, x ≈ 2.58 (rounded to three significant figures).
So, the two solutions to the equation are x = 3 and x ≈ 2.58.