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[Arithela]

simplify: e^3ln3


can someone please go through it step by step with the laws?

 

[Trebla]

e^3ln3 = e^{ln 33}
(remember that: ln a^x = xln a)
= 3³
(remember that e^{ln x} = x, i.e. a function of its own inverse gives the variable)
= 27
You can also verify it on your calculator...

 

[Arithela]

thanks! but i did not know that e^lnx = x


why is this rule not in textbooks?

 

[lyounamu]

thanks! but i did not know that e^lnx = x


why is this rule not in textbooks?

It's not in textbook because it is a question where you just apply your knowledge of the log rule to the real question.

For example, we don't actually learn that tan^2(x) = sec^2(x) - 1. But we derive this rule from the rule that is in the textbook (which is tan^2(x) + 1 = sec^2(x)).

 

[3unitz]

thanks! but i did not know that e^lnx = x


why is this rule not in textbooks?

let y = e^ln x

take log of both sides:

ln y = ln (e^ln x)

ln y = ln x . ln e (using ln (x^a) = a ln x)

ln y = ln x

.'. y = x

hence x = e^ln x

It's not in textbook because it is a question where you just apply your knowledge of the log rule to the real question.

For example, we don't actually learn that tan^2(x) = sec^2(x) - 1. But we derive this rule from the rule that is in the textbook (which is tan^2(x) + 1 = sec^2(x)).

thats derived from sin²x + cos²x = 1 by dividing both sides by cos²x

 

[lyounamu]

thats derived from sin²x + cos²x = 1 by deviding both sides by cos²x

I know. But in the textbook, they give that as a "formula". Well, it doesn't matter.