limit, x->2, of ln(e^(ln(5x))); limit, x->2, ln(e^(5 ln(x)))

[uniqueownz]

Find each of the following limits.

lim ln e[sup:2l6rs4rn]ln 5x[/sup:2l6rs4rn]
[sup:2l6rs4rn]x->2[/sup:2l6rs4rn]

lim ln e[sup:2l6rs4rn]5 ln x[/sup:2l6rs4rn]
[sup:2l6rs4rn]x->2[/sup:2l6rs4rn]

 

[skeeter]

Re: LIMITS URGENT!!!

hint ...

\(\displaystyle e^{\ln(a)} = a\)

 

[stapel]

lim ln e[sup:3dngprlh]ln 5x[/sup:3dngprlh]
[sup:3dngprlh]x->2[/sup:3dngprlh]

What is the value of 5x for x = 2? What then is the value of e[sup:3dngprlh]ln(5x)[/sup:3dngprlh]? What then is the value of ln(e[sup:3dngprlh]ln(5x)[/sup:3dngprlh])?

lim ln e[sup:3dngprlh]5 ln x[/sup:3dngprlh]
[sup:3dngprlh]x->2[/sup:3dngprlh]

Use a log rule to convert 5 ln(2) into an equivalent log expression. Then evaluate e raised to that power. Then take the log.



Eliz.

 

[uniqueownz]

thank you both for responding but im still confused on both problems :/

like ok e[sup:22twwvsc]ln 5x[/sup:22twwvsc] = 5x and x = 2 therefore, it equals 10. then i have to find ln 10. how do i do this?

and for the second one, is the log rule that: 5 ln x = ln x[sup:22twwvsc]5[/sup:22twwvsc] ? if so that would mean i have to find ln 32. how do i do this?

 

[skeeter]

thank you both for responding but im still confused on both problems :/

like ok e[sup:aa6npm15]ln 5x[/sup:aa6npm15] = 5x and x = 2 therefore, it equals 10. then i have to find ln 10. how do i do this?

and for the second one, is the log rule that: 5 ln x = ln x[sup:aa6npm15]5[/sup:aa6npm15] ? if so that would mean i have to find ln 32. how do i do this?

no one expects you to know or find the value of ln(10) or ln(32), they're both just constants. get out a calculator if you want to see their decimal approximation.

 

[uniqueownz]

mhmmmm negative.
how does that work :/