e^2x-1 = 5 + k

[kinkywink]

hey guys, just wondering if im doing this question the right way...

 

i went from e^(2x-1) = 5 + k to e^(2x-1) -5= k

then to lne^(2x-1) -5 = lnk

 

which simplifies to (2x-1) -5 = lnk

 

does then then go to 2x-6 =lnk?

 

then 2x = lnk +6

 to x = (lnk +6)/2   ?

[MegaDave91]

What? Engrish prease.

[kinkywink]

What? Engrish prease.

do you even math

[MegaDave91]

do you even math

Quite terrible at it.

[kinkywink]

Quite terrible at it.

yeah.. me too

[WorldDominator]

Looks right to me. (Math major)

 

Edit: wait you didnt do ln(5)

[DirtyFish]

Aren't you supposed to ln the whole thing? So you should have to have ln(5) in there somewhere?

[helping]

e^2x-1 = 5 + k -> e^2x = 6+k

lne^2x = ln(6+k)

x = (ln(6+k)/2

 

you should be good.

 

(I'm assuming you mean e^2x-1 and not e^2x-1)

otherwise you would have to add 1 after taking the natural log. 

[othertomperson]

You need to use brackets more consistently.

 

e^2x-1 = 5 + k

=> e^2x=6+k

=> 2x=ln(6+k)

=> x= 1/2 * ln(k+6)

but

e^(2x-1) = 5+k

=> 2x-1  = ln(5+k)

=> x = 1/2 * (ln(5+k) +1)

 

 

I know some people prefer to use the notation Exp[x+y]+z and Ln[x+y]+z because it's more obvious what you're talking about.

[kinkywink]

e^2x-1 = 5 + k -> e^2x = 6+k

lne^2x = ln(6+k)

x = (ln(6+k)/2

 

you should be good.

 

(I'm assuming you mean e^2x-1 and not e^2x-1)

 the e^2x-1 is bracketed ? howd you get it to ln (6+k)

[kinkywink]

Um did you add the brackets to the second line yourself, or did you just omit them from the first line? Because they aren't the same thing.

yeah that was my bad while typing

[kinkywink]

Looks right to me. (Math major)

 

Edit: wait you didnt do ln(5)

ah, i didnt know if i was meant to do that

[helping]

 the e^2x-1 is bracketed ? howd you get it to ln (6+k)

I interpreted that as e^2x-1 = 5+k

 

since it's e^2x-1 = 5 + k then you would do 2x-1 = ln(5+k) --> x = (ln(5+k)+1)/2 

[AlwaysFSX]

e^2x-1=5+k

(e^2x-1)-5=k

(2x-1)-5=ln[k]

2x-6=ln[k]

x=(ln[k]+6)/2

 

Unless we both did it wrong? I haven't done this in a while

[kinkywink]

I interpreted that as e^2x-1 = 5+k

 

since it's e^2x-1 = 5 + k then you would do 2x-1 = ln(5+k) --> x = (ln(5+k)+1)/2 

cheers i was just thinking this! thanks

 

e^2x-1=5+k

(e^2x-1)-5=k

(2x-1)-5=ln[k]

2x-6=ln[k]

x=(ln[k]+6)/2

 

Unless we both did it wrong? I haven't done this in a while

k+5 needs to be lned and bracketed

[AlwaysFSX]

cheers i was just thinking this! thanks

 

k+5 needs to be lned and bracketed

Son of a... I forgot about that.

[Vitalius]

yeah that was my bad while typing

Guys. Guys. Guys.

^{There's a super script option...}_{And a sub script option.} Do you even format?

Just saying, this is the exact situation those things are made for.

[helping]

Guys. Guys. Guys.

^{There's a super script option...}_{And a sub script option.} Do you even format?

Just saying, this is the exact situation those things are made for.

Both are for _p^le^bs 

 

real men just yolo e^allthevariableshere+5

 

(I just noticed we had these when I needed them earlier)

[othertomperson]

Looks right to me. (Math major)

 

Edit: wait you didnt do ln(5)

 

He did, he just wrote it outside the bracket. The next line he's simplified it as though it were inside the log.

[CatCloud]

I came here soloy to say i have noooo clue on wtf all this is or is good for!

Keep it up!

[DoubleY]

I came here soloy to say i have noooo clue on wtf all this is or is good for!

Keep it up!

[Askew]

[CatCloud]

https://www.youtube.com/watch?v=bX7V6FAoTLc

Sorry due to lame copyright reasons, i cant watch that.

[DoubleY]

Sorry due to lame copyright reasons, i cant watch that.

I cri eritime.

[CjlabanG]

e^x=y is lny=x  

 

you could just:

 

e^(2x-1) = 5+k

 

ln(5+k)=2x-1        

 

ln(5+k)+1 = x

      2

 

Simple as that.