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imhimanshu
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AbhiJ

Basically you are saying 2 raised to some irrational value = 1. Any irrational should lie between -infinity and plus infinity.

see the graph of a^x

https://www.mathnotes.org/content/math/f ... _graph.gif
Hi AbhiJ,

I didn't get your point over here. Could you please explain it in detail.

Thanks
imhimanshu
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In the graph of \(y = a^x\) you can see at only one point the value of the y = 1. That is at x = 0. At no other value between - infinity and + infinity the value of y becomes 1. It's an always increasing curve. Irrational value also lie between - infinity and + infinity . Hence at no other irrational value is y = 1.
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imhimanshu
Hi Instructors,
Request you to please clarify the below doubt on the usage of the property

\(x^y\) = 1

If X is not equal to 0 and\(x > 1\); Will the value of y be always 0.
I mean, isnt it possible that y has some irrational value that makes \(x^y\) = 1

Example; \(2^(1/irrational value) = 1\). If not, what can we infer about the value of whole expression if \(x^(1/irrational value)\)

If \(x^y = 1\) and \(x>1\) then \(y=0\) only. For example, \(2^y=1\) --> \(y=0\).

Hope it's clear.
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imhimanshu
Hi Instructors,
Request you to please clarify the below doubt on the usage of the property

\(x^y\) = 1

If X is not equal to 0 and\(x > 1\); Will the value of y be always 0.
I mean, isnt it possible that y has some irrational value that makes \(x^y\) = 1

Example; \(2^(1/irrational value) = 1\). If not, what can we infer about the value of whole expression if \(x^(1/irrational value)\)


Responding to a pm:

You can also do another thing to understand this. Solve:

\(2^{1/x} = 1\)

Taking log on both sides

\((1/x) log 2 = log 1\)
\((.3/x) = 0\)

Is there a real value for x?
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imhimanshu
Hi Instructors,
Request you to please clarify the below doubt on the usage of the property

\(x^y\) = 1

If X is not equal to 0 and\(x > 1\); Will the value of y be always 0.
I mean, isnt it possible that y has some irrational value that makes \(x^y\) = 1

Example; \(2^(1/irrational value) = 1\). If not, what can we infer about the value of whole expression if \(x^(1/irrational value)\)

A number that can be expressed as a ratio of two numbers is rational. However, roots, \(\pi\), and endless non-repeating decimals are irrational numbers; they cannot be written in a ratio of two numbers. Moreover, an infinite amount of irrational numbers exist between integers. But, in no way, there is an infinite amount of irrational numbers that make integers, applying the logic that you used above.

I believe you are thinking as if rational and irrational numbers acted like odd and even numbers. Even though O + O = E or O + E = O, no such relationship exists between rational and irrational numbers.

I hope this helps.



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