Question

Hello i am having trouble understanding what a logarithm is specifically, any help would be greatly appreciated

Answer 1

4 and4 is 8, right?

Answer 2

4 = 2^2 and 8 = 2^3, agree?

Answer 3

yes

Answer 4

and 16 is 2^4

Answer 5

agreed

Answer 6

2^2 * 2^2 = 2^4, yes?

Answer 7

yes

Answer 8

What would 2^2 *2^2*2^2 be?

Answer 9

2^6

Answer 10

Which u got by adding the exponents, right?

Answer 11

yes

Answer 12

That's what a logarithm does, converts multiplication into addition.

Answer 13

In this case u would say the base is 2.

Answer 14

ok, that somewhat clarifys it, thank you

Answer 15

Somewhat?

Answer 16

its just that i dont understand WHAT a logrithim is, what you told me was how a logrithim is a base just as it is the exponents being added togeather

Answer 17

log to base a of y MEANS y = a^x

Answer 18

Is that clearer?

Answer 19

yes, much

Answer 20

x = log_a y is same as y = a^x

Answer 21

4 = log_2 16 and 16 = 2^4

Answer 22

Sure u got it?

Answer 23

When the base is e, you say ln instead of log and call them "natural logarithms"

Answer 24

ok! wow that really hlped thank you!

Answer 25

if u were to ask the question. 2 to cubed equals what? u would have 2^3=y. y being what we are solving for.
so
2^3=2*2*2=8 so the answer we are looking for is 8
from this we can get a general equation for values of 2 as we change the exponent
2^x=y
so we change x to say 3 0 or 4 and we solve
2^4=16
so generally we are asking the question a number(like 2) to a power that we cahnge(or the x variable) equals what(the y variable). or
a^x=y

Now for logs we are asking a different question. suppose i said 8 is 2 to what power?
then u would have
8=2^y and we put y here because it is what we are solving for
of course we know this is 3 since we did it earlier
but suppose i asked the question:
now from this we can get a general equation to solve
x=2^y
as we change x we solve for the y values
the even more general statement would b
a number to what power equals the number we change or mathematically
x=a^y

now we take this equation and solve for y because we like it on 1 side by itself. but there is a problem here. how do we get the y by itself? so what we do is we create a log system which the opposite of the exponential system. thus all the rules of exponents and logs are reversed.

so what we do is we take the log of each side. the base being a (or the constant)
log(base a)(x)=log(base a)a^y
log(base a)(x)=y

now the important thing here is that
log(base a)(x)=y
and
x=a^y
is the same exact equation and asks the same exact question