E^(2lnt) = t^2 Whyyy?

[CinderBlockFist]

ok guys, i don't see how e^2ln|t| = t^2 can someone explain it to me please? Seems so easy but i don't see it.

 

[CinderBlockFist]

ok, so far i tried this. I know e^(lnx) = x


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)

 

[Brad Barker]

ok, so far i tried this. I know e^(lnx) = x


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.

 

[quantumdude]

First look at this rule for logarithms:

$\log_b(a^x)=x\log_b(a)$.

Now apply that to your exponent:$2\ln |t|$. What do you get?

Then note that $f(x)=\ln (x)$ and $g(x)=e^x$ are inverse functions, which means that $f(g(x))=g(f(x))=x$.

Those two rules together will give you the answer.

 

[CinderBlockFist]

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