simple concept which i can not understand (1 Viewer)

[girlworld_club]

How does:

1. e^20k = 2 equal
20k = ln2

what is the working in between?

2. also what is e^x = 0

3. and another one, how would you solve e^x = 20
i.e what is x?

4. Also how to solve 2^x


thank you so much for your time.

 

[HeroicPandas]

1.

Imagine a hammer smashing down the "20k". Where can the e go? It goes to another universe, transforms into an "ln or log base e" and attaches itself to that lonely 2

2. e^x = 0

x = ln(0) = impossibru

U can draw out the curve y = e^x. Notice that y > 0? Therefore y =/= 0, therefore e^x =/= 0

3. 2^x? Impossibru again....how can u solve something that has no LHS and RHS

Do u mean differentiate? Integrate?

If so, please recall a special rule:

 

[asianese]

For the first one - reacall the definition of a logarithm

. You can use this definition directly to show that

 

[asianese]

1.

Imagine a hammer smashing down the "20k". Where can the e go? It goes to another universe and attaches itself to that lonely 2

No, the e NEVER goes into another universe! All parts of that equation stay there, just in a different form. The e becomes the base of the logarithm...without a base a logarithm is meaningless. Obviously this is what you tried to imply lol

 

[HeroicPandas]

no, the e never goes into another universe! All parts of that equation stay there, just in a different form. The e becomes the base of the logarithm...without a base a logarithm is meaningless. Obviously this is what you tried to imply lol

hahahahaha!

 

[RishBonjour]

How does:


4. Also how to solve 2^x

As HeroicPandas pointed it, what do you mean? it is a typical question in 2u so I assume differentiate?
If thats the case, there is a 'formula' (just a shortcut) can't seem to remember, but this is how I would do it:

y = 2^x

lets make x the subject so we can differentiate normally:

lny = ln(2^x)

therefore, lny = x ln2
rearrange --> x = (lny)/(ln2)

this can now be easily differentiated (with respect to y)

dx/dy = 1/(yln2)

therefore, dy/dx = (yln2)
since y = 2^x ---> dy/dx = (2^x)(ln2)


if integrating, you could think of it as 'what can I differentiate to get 2^x' (usually the question will guide you with this).
well it would be '(2^x)/(ln2) + C'

1.

Imagine a hammer smashing down the "20k". Where can the e go? It goes to another universe and attaches itself to that lonely 2 by fusing with a log first

Fuaaarrk

 

[Parvee]

1.

Imagine a hammer smashing down the "20k". Where can the e go? It goes to another universe and attaches itself to that lonely 2 by fusing with a log first

damnnnnn

 

[girlworld_club]

Also how do we integrate In2

??? thanks

 

[SpiralFlex]