Question

Can someone explain to me why e^ln 7 is equal to 7?

Answer 1

e and ln are inverses of each other

Answer 2

Think of an analogy: why does 7 divided by 9 and multiplied by 9 equal 7?

Similarly, ln 7 means "e raised to what power is 7?". Then raise e to that power, and you get 7.

Answer 3

Why isn't it 1 to the seventh power?

Answer 4

"ln 7" is the value that, when you raise e to it, gives you 7. "e to what power is 7?" is ln 7.

Answer 5

Because a log is an exponent. So it must be placed on a base. when you write ln 7 you are really saying tell me the exponent that I must put on the base e so that I will get 7. So then you place it on the base e. And WALLA!!! you get 7

Answer 6

*voila

Answer 7

I still don't understand.

Answer 8

Let's use a different base, like 2. What's log (base 2) of 16?

Answer 9

In other words, 2 to what power equals 16?

Answer 10

\[\text{ let } y=e^{\ln(7)}\]
we could take natural log of both sides
\[\ln(y)=\ln(e^{\ln(7)})\]
using properties of log we can rewrite this
\[\ln(y)=\ln(7) \cdot \ln(e)\]
but ln(e)=1
so we have
\[\ln(y)=\ln(7)\]
but this would imply y=7

Answer 11

I get that but i still don't understand the e ln thing

Answer 12

Do you see why 2 ^ (log base 2 of 16) equals 16?

Answer 13

@ktklown, yes
@myininaya, what is 'e'? and I don't understand how the equation was rewritten. Wouldn't it be ln^2(7) times ln(e)?

Answer 14

e is a number that is used a lot in calculus and some other classes. It is about 2.7 So when you write ln what you are really writing is log(base 2.7)

Answer 15

Confucious -- ok, so if you see why 2 ^ (log base 2 of 16) is equal to 16, then, you can probably also see why 2.718 ^ (log base 2.718 of 16) is equal to 16.

"e" is just a special number, about 2.71828, and "ln" just means "log base e".

Answer 16

\[\ln 7 =\log_{e}7 \]

Answer 17

When I put log(base 2.7) 7 in the calculator I don't get 7. I don't get that when I put in log(base e) 7 either.

Answer 18

you shouldn't get 7 if you put that

Answer 19

How did yo rewrite the equation?

Answer 20

No, but 2.7 to the POWER of log (base 2.7) of 7 is equal to 7.

Answer 21

you have
\[e^{\ln(7)}=e^{\log_e(7)}=7\]

Answer 22

log (base 2.7) of 7 is not 7, it's the number that, if you raise 2.7 to that power, will give you 7.