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The value of 8^x is always greater than which number?

0,1,8
please only pick one and brainless

User Panayotis
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2 Answers

4 votes

Explanation:

If x is positive or negative 8^x will always be greater than 0

examples 8^0 = 1

8^-4 = 1/8^4 = .0002441 still greater than zero (but less than one)

User Vincent J
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For any real number x, 8^x is always greater than 0 due to the exponential nature of the function, ensuring positivity even as x approaches negative infinity.

The value of 8 raised to the power of x, denoted as 8^x, is always greater than 0. This holds true for any real number x, except in the case when x tends towards negative infinity, where 8^x approaches zero but never equals it.

When x is positive, 8^x increases exponentially, ensuring that the result is always a positive value. Even when x is 0, 8^x equals 1, indicating that it is greater than 0. As x becomes negative but remains a real number, 8^x approaches 0 but never quite reaches it.

This behavior is a fundamental property of exponentiation, where the base (in this case, 8) raised to any real power always yields a positive result. Therefore, we can confidently state that the value of 8^x is consistently greater than 0 for all real numbers x.

User JohnDotOwl
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