Can e^3 be converted to a logarithmic function with a value of 0.0498?

[shoook]

1. Convert to logarithmic function: e^3=0.0498
3. ln(-3)=0.0166 is my guess

 

[rocomath]

you cannot take the ln of a negative number

 

[shoook]

okay well I was wrong, could it be ln(3)=0.0166?

 

[rocomath]

where is .01666 coming from?

 

[shoook]

i divided 0.0498 by 3. that is totally wrong though.

 

[rocomath]

i divided 0.0498 by 3. that is totally wrong though.

yes.

since it just asks you to convert to log form, just leave it

$$e^{3}=0.0498$$

$$ln(e^{3})=ln(0.0498)$$

 

[shoook]

okay if the equation was e^-3=0.0498 would I just rewrite it as:
ln(e^-3)=ln(0.0498) ? Also, does ln remain on the right side as well as the left?

 

[rocomath]

no they cancel each other out, i was just showing you the operation; sorry for the confusion.

$$3=ln(0.0498)$$

 

[Dick]

Uh, you mean -3=ln(0.0498...), right?

 

[rocomath]

okay if the equation was e^-3=0.0498 would I just rewrite it as:
ln(e^-3)=ln(0.0498) ? Also, does ln remain on the right side as well as the left?

then it would be

$$-3=ln(0.0498)$$

 

[rocomath]

Uh, you mean -3=ln(0.0498...), right?

that's his 2nd question

 

[Dick]

Well, ok, however you want to handle it. But seeing stuff like 3=ln(0.0498) makes me nervous.

 

[rocomath]

Well, ok, however you want to handle it. But seeing stuff like 3=ln(0.0498) makes me nervous.

lol sorry!