- #1

CinderBlockFist

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- Thread starter CinderBlockFist
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- #1

CinderBlockFist

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- #2

CinderBlockFist

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so, i broke the e^(2ln|t|) into to parts:

e^2 times e^(ln|t|) which equals t (from the top identity)

so I am left with e^2 times t. which is te^2

but the book says it equals t^2..so what happened to the e? (exponential function)

- #3

Brad Barker

- 429

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CinderBlockFist said:

so, i broke the e^(2ln|t|) into to parts:

e^2 times e^(ln|t|) which equals t (from the top identity)

so I am left with e^2 times t. which is te^2

but the book says it equals t^2..so what happened to the e? (exponential function)

you have

e^2 e^(ln t) = e^(2+ln t),

which is incorrect.

you want to use the properties:

a ln b = ln b^a

and

e^ln a = a.

the rest should be straightfoward.

Last edited:

- #4

quantumdude

Staff Emeritus

Science Advisor

Gold Member

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[itex]\log_b(a^x)=x\log_b(a)[/itex].

Now apply that to your exponent:[itex]2\ln |t|[/itex]. What do you get?

Then note that [itex]f(x)=\ln (x)[/itex] and [itex]g(x)=e^x[/itex] are

Those two rules together will give you the answer.

- #5

CinderBlockFist

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SWEEET! THANK YOU GUYS. I got my laws of exponents mixed up. :yuck:

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