Everything Power Zero Is One, Right?

Algebra Level 3

True or false :

\quad \quad log x 1 = 0 \log_{x}1=0 for every x > 0 x>0 .

True False

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3 solutions

This is false for x = 1 x=1 .

The matter is that using 1 0 = 1 1^0=1 is the same that 1 4 = 1 1^4=1 , that is the reason why log 1 1 0 \log_{1}1\ne 0 , because by that logic log 1 1 = 4 \log_{1}1=4 is true too and it means log 1 1 = log 1 1 0 = 4 \log_{1}1=\log_{1}1 \Rightarrow 0=4 , which is false...... I hope this explains what you want to know, feel free of asking whatever you want

I konw x=1 is not in Domain of log! But 1°=1 . May you explain more?

Mohamad Zare - 5 years, 3 months ago

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The matter is that using 1 0 = 1 1^0=1 is the same that 1 4 = 1 1^4=1 , that is the reason why log 1 1 0 \log_{1}1\ne 0 , because by that logic log 1 1 = 4 \log_{1}1=4 is true too and it means log 1 1 = log 1 1 0 = 4 \log_{1}1=\log_{1}1 \Rightarrow 0=4 , which is false...... I hope this explains what you want to know, feel free of asking whatever you want

Hjalmar Orellana Soto - 5 years, 3 months ago

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It was useful! Thank you!

Mohamad Zare - 5 years, 3 months ago

True! That got me!

Colin Carmody - 5 years, 3 months ago

The name of the problem gives away the answer though.

Arihant Samar - 5 years, 3 months ago

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I left this problem as level 2 at the beginning, the idea with a name which make the answer sort of obvious is confusing, there are a lot of ones which think that is impossible to make a problem with the answer in the name, also, in brilliant all we are supposed to be here for making and solving problems, it has no sense to be guessing only because the name says something, and I guess that @Pi Han Goh , thinks similar about his problem

Hjalmar Orellana Soto - 5 years, 3 months ago

As far as I know, logarithm of base 1 is not defined (indeterminate), so I am afraid your explanation cannot convince me.

Desmond Yeung - 5 years, 3 months ago

but by definition of log, logarithm to the base 1 is never defined,there is a restriction that the base can never be equal to 1....

Shubh Karman Singh - 4 years, 10 months ago

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That's just why the statement is false, 1 > 0 1>0 and we can't take x = 1 x=1 ....

Hjalmar Orellana Soto - 4 years, 10 months ago
Varun Shekhar
Apr 14, 2018

logarithmic functions are defined only when x > 0 x>0 and x x not equal to 1.

Sorry I don't understand, I always thought that 1 0 = 1 1^{0} = 1 , is this wrong? Why?

We know that 1 n = 1 { 1 }^{ n }=1 is true for all n n (except for \infty ). Let me take two examples: 1 0 = 1 { 1 }^{ 0 }=1 and 1 1 = 1 { 1 }^{ 1 }=1 . But, log 1 1 0 \log _{ 1 }{ 1 } \neq 0 and log 1 1 1 \log _{ 1 }{ 1 } \neq 1 .

Suppose log 1 1 = 0 \log _{ 1 }{ 1 } =0 and log 1 1 = 1 \log _{ 1 }{ 1 } =1 are true, then

log 1 1 = log 1 1 0 = 1 \log _{ 1 }{ 1 } =\log _{ 1 }{ 1 } \\ \Rightarrow 0=1 .

There is a contradiction. Therefore, log 1 1 \log _{ 1 }{ 1 } is undefined. We can then conclude that log x 1 \log _{ x }{ 1 } is true when x 1 x\neq 1 .

Kenneth Choo - 5 years, 3 months ago

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I LOVE YOUR ANSWER

Nicko Valero - 3 years, 7 months ago

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