Question about why ln(e^x) =/= x

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Discussion Overview

The discussion revolves around the equation ln(e^x) and why it is perceived as not equal to x. Participants explore the implications of inverse functions, domain and range considerations, and comparisons with other mathematical functions.

Discussion Character

  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant questions why ln(e^x) is not equal to x, noting that the domain and range of both sides appear to be the same.
  • Another participant asserts that ln(e^x) is indeed equal to x, providing a mathematical simplification involving ln(e).
  • A third participant agrees with the assertion of equality but expresses confusion based on previous information from Wolfram Alpha, which they believed indicated the equation was false.
  • A different participant highlights that the equation e^(ln x) = x is only valid for x > 0, suggesting that domain considerations are crucial in understanding these relationships.

Areas of Agreement / Disagreement

There is disagreement among participants regarding the equality of ln(e^x) and x, with some asserting it is true and others questioning this conclusion based on domain issues.

Contextual Notes

Participants mention domain restrictions, particularly in relation to the logarithmic function, which may affect the validity of certain equations.

physicsernaw
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Why is ln(e^x) =/= x?

The domain and range of the LHS are the same as the RHS, so I don't understand why this equation is false, where e^ln(x) = x, and the LHS and RHS of this does not have the same domain...

I know that e^x and ln(x) are inverse functions, so please don't only tell me this. Why does e^ln(x) = x, while ln(e^x) =/= x?

EDIT:
Like, I understand why sin^-1(sin(x)) =/= x, whereas sin(sin^-1(x)) = x, and this is because sin^-1(x) has a range of -pi/2 to pi/2
 
Last edited:
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They are equal.

ln(e^x) = x*ln(e) = x

What/who told you differently?
 
Vorde said:
They are equal.

ln(e^x) = x*ln(e) = x

What/who told you differently?

I could have sworn wolfram alpha was telling me the equation is false, but I just tried again and it told me the equation is indeed true :blushing:
 
It's the other way around that you run into domain considerations:

$$ e^{\ln x} = x$$

The left side is defined only for x > 0 (considering only the real-valued log function).
 

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