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Find all the values of 𝑥 which satisfy 7^(𝑥(𝑥 + 2)) = 1.
Find all the values of 𝑥 which satisfy seven to the power of 𝑥 multiplied by 𝑥 plus two is equal to one.
In order to answer this question, we recall that any nonzero number raised to the power of zero is equal to one. This means that seven to the power of zero is equal to one. We can then rewrite the right-hand side of our equation as seven to the power of zero such that seven to the power of 𝑥 multiplied by 𝑥 plus two is equal to seven to the power of zero.
At this stage, both sides of our equation have a base of seven. This means that the exponents must be equal. 𝑥 multiplied by 𝑥 plus two is equal to zero. As the product of 𝑥 and 𝑥 plus two equals zero, then either 𝑥 equals zero or 𝑥 plus two equals zero. We can solve this second equation by subtracting two from both sides such that 𝑥 is equal to negative two.
The values of 𝑥 which satisfy the equation seven to the power of 𝑥 multiplied by 𝑥 plus two equals one are negative two and zero.
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