Question

What is the domain of y=ln(x+3)+2?
also how do i find the domain of any logarithmic function in general.

Answer 1

you can only take a log of something that is greater than 0, and therefore `x+3` has to be greater than 0.

x+3>0

now, solve for x./

Answer 2

and that will give me the domain?

Answer 3

yes, if you solve it correctly

Answer 4

I get x>-3

Answer 5

yes

Answer 6

that means that x can be any number greater than -3.

Answer 7

Okay thanks. is that it? so its just all numbers bigger than -3?

Answer 8

in this case, yes/

Answer 9

you want a little general review too, or not?

Answer 10

Sure:)

Answer 11

Also can you help me to find the range???

Answer 12

The domain of a logarithmic function in general:

\(\large\color{black}{ \displaystyle {\rm y} =\log_{\rm a}\left(~ {\rm f}(x)~\right)}\)

above \(\color{blue}{\LARGE\Uparrow}\) is a logarithmic function.
(it is a y, function of x. It has some positive base a, and any involving x function f(x) the log of which you are taking.)

in this case the domain is the x-solutions that make f(x) be greater than 0.
(that is pretty much all)

Answer 13

however the range would be \((-\infty,~+\infty)\)

(this is also a good place to introduce the concept of a "limit" )

Answer 14

Okay well how do you find the range using the equation that I had?

Answer 15

have you previously been finding the range of such (logarithmic) functions?

Answer 16

I will start from a pretty simple example

Answer 17

I think I may have done things like this last year when I was in algebra 2 but now i am in precalc and I dont remember how to find the range or domain of a logarithmic function.

Answer 18

\(\large\color{black}{ \displaystyle {\rm y}=\log(x) }\)
(when the base of a log is not specified, the rule is that this base is 10)

Obviously, as x approaches \(\infty\) (saying - gets infinitely bigger and bigger), the output [in this case] is also getting larger and larger.

So, we can tell that as \(x\rightarrow \infty\), then \(y\rightarrow \infty\).

Answer 19

so, are you good so far with why when x increases infinitely the y (the output) will also grow towards infinity?

Answer 20

we can test this by plugging in values too;
\(\large\color{black}{ \displaystyle {\rm y}=\log_{10}(x) }\)

\(\large\color{black}{ \displaystyle (10,1) }\)
\(\large\color{black}{ \displaystyle (100,2) }\)
\(\large\color{black}{ \displaystyle (1000,3) }\)
and so on....
\(\large\color{black}{ \displaystyle (10^{\rm n},{\rm n}) }\)

this is growing very slowly, but it will and certain can get an infinitely large output.

Answer 21

ask anything so far, or verify that you got everything so far. I don't want to go on without that.

Answer 22

I get it so far

Answer 23

@SolomonZelman

Answer 24

Ok,

Answer 25

Now, lets see what happens to \(\large\color{black}{ \displaystyle {\rm y}=\log(x) }\) when the value of x we are plugging in closer and closer approaches 0.

Answer 26

What is the value of the function when x=0.1?

\(\large\color{black}{ \displaystyle y= \log(x) }\)

plugging 0.1 for x

\(\large\color{black}{ \displaystyle y= \log(0.1) }\)

re-writing as a fraction

\(\large\color{black}{ \displaystyle y= \log\left(\frac{1}{10}\right) }\)

now I will apply some rules that exponents have.

\(\large\color{black}{ \displaystyle y= \log\left(\frac{1}{10^{1}}\right) }\)


\(\large\color{black}{ \displaystyle y= \log\left(10^{-1}\right) }\)

the exponent of the log, goes on the outside (you know that, right?)

\(\large\color{black}{ \displaystyle y= (-1)\log\left(10\right) }\)

log(10) is 1, so

\(\large\color{black}{ \displaystyle y= (-1)\times 1=-1 }\)

so when \(x=0.1\) then \(y=-1\)

Answer 27

while you read this, I will post some more values, to give you an idea of the function behavior.

Answer 28

Okay got it

Answer 29

I knew that the exponent of the log goes on the outside before the log

Answer 30

and I learned about how 10 is the number you use as a base if there is not one given with general logs and with natural logs you use e

Answer 31

We have a point (0.1, -1). NOW, we will plug some x values that are closer to 0, and will follow the same algorithm (the same steps) to find y.

what is the v-value at x=0.01? (choose 0.01 that is closer to 0, and plugged it into the function)

\(\large\color{black}{ \displaystyle y= \log(0.01) }\)

\(\large\color{black}{ \displaystyle y= \log\left(\frac{1}{100}\right) }\)

\(\large\color{black}{ \displaystyle y= \log\left(\frac{1}{10^{2}}\right) }\)

\(\large\color{black}{ \displaystyle y= \log\left(10^{-2}\right) }\)

\(\large\color{black}{ \displaystyle y= (-2)\log\left(10\right) }\)

\(\large\color{black}{ \displaystyle y= (-2)\times 1=-2 }\)

so when \(x=0.01\) then \(y=-2\)
so the points as x approaches zero so far go the following way:
(0.1 , -1)
(0.01 , -2)

Answer 32

yes, glad you are getting it, and what you said is correct;)

Answer 33

I will show more values

Answer 34

when x=0.001, y=?

\(\large\color{black}{ \displaystyle y= \log(0.001) }\)

\(\large\color{black}{ \displaystyle y= \log\left(\frac{1}{1000}\right) }\)

\(\large\color{black}{ \displaystyle y= \log\left(\frac{1}{10^{3}}\right) }\)

\(\large\color{black}{ \displaystyle y= \log\left(10^{-3}\right) }\)

\(\large\color{black}{ \displaystyle y= (-3)\log\left(10\right) }\)

\(\large\color{black}{ \displaystyle y= (-3)\times 1=-3 }\)

so when \(x=0.001\) then \(y=-3\)
so the points as x approaches zero so far go the following way:
(0.1 , -1)
(0.01 , -2)
(0.001 , -3)

Answer 35

and so would be true for x=n (for any negative integer n)

Answer 36

Okay

Answer 37

\(\large\color{black}{ \displaystyle y= \log(0.\underline{0...0}1) }\)

(n zeros ^)

\(\large\color{black}{ \displaystyle y= \log\left(\frac{1}{1\underline{0...0}}\right) }\)

(n zeros ^)

\(\large\color{black}{ \displaystyle y= \log\left(\frac{1}{10^{n}}\right) }\)

\(\large\color{black}{ \displaystyle y= \log\left(10^{-n}\right) }\)

\(\large\color{black}{ \displaystyle y= (-n)\log\left(10\right) }\)

\(\large\color{black}{ \displaystyle y= (-n)\times 1=-3 }\)

so when \(x=0.\underline{0...0}1\) then \(y=-n\)

(n zeros ^)

so the points as x approaches zero so far go the following way:
(0.1 , -1)
(0.01 , -2)
(0.001 , -3)

... etc, and till

\((0.\underline{00...0}1~,~-n)\)

(n zeros ^)

Answer 38

just as you can plug a super-close to 0 x-value, when x= (some tiny decimal),
same way you can get a negative with a huge magnitude.

Answer 39

So really its just (-infinity, +inifinity) as the range?

Answer 40

what about finding the x Intercept??

Answer 41

x intercept:
set y=0, and solve for x

y intercept:
set x=0, and solve for y

Answer 42

well, y-intercept wouldn't always exist .... but if it does, than this is what you got to do.