Part A: If (7^2)^x = 1, we can simplify the expression by applying the exponent rule which states that (a^b)^c = a^(b*c). Therefore, we have 7^(2*x) = 1.
In order for an exponentiation to result in 1, the base must be 1, since any non-zero number raised to the power of 0 is equal to 1.
Therefore, we have 7^(2*x) = 1, which means 7^(2*x) = 7^0. By equating the exponents, we can conclude that 2*x = 0.
Solving for x, we divide both sides of the equation by 2, yielding x = 0/2, which simplifies to x = 0.
So the value of x in the equation is 0.
Part B: If (7^0)^x = 1, we can simplify the expression by again applying the exponent rule, which states that a^0 = 1, where a is any non-zero number.
Therefore, we have 1^x = 1.
Since any number raised to the power of 1 is itself, we can conclude that x must be any real number.
Therefore, the possible values of x are all real numbers.
Part A: If (7^2)^x = 1, what is the value of x? Explain your answer. (5 points)
Part B: If (7^0)^x = 1, what are the possible values of x? Explain your answer. (5 points)
1 answer