Graph y=log 3(x+2)-1. Then, find the domain/range

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- anonymous

Graph y=log 3(x+2)-1. Then, find the domain/range

- jamiebookeater

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- anonymous

what does "solve" mean in this context?

- anonymous

Whoops! Not solve. Just graph and find the domain/range

- jim_thompson5910

I would use a graphing calculator like desmos
https://www.desmos.com/calculator
to graph this out. You will have to use the change of base formula to make sure the expression is typed in properly
you will have to type in log(x+2)/log(3) - 1

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## More answers

- anonymous

Ok thanks :) Would the domain be all real numbers?

- jim_thompson5910

why all real numbers?

- jim_thompson5910

you should have this graph
https://www.desmos.com/calculator/fs1pnhr9hx

- jim_thompson5910

the left side of the graph seems to hit a "wall" of some sort

- anonymous

Ok, now I do. And it does seem like that

- jim_thompson5910

it turns out that you cannot plug in 0 or negative numbers into a log function
so x+2 has to be positive
x+2 > 0 leads to x > ???

- anonymous

x=-2 ? x+2=0

- jim_thompson5910

yeah x = -2 is your vertical asymptote. The graph does NOT touch the vertical asymptote. It only gets closer and closer and closer.

- jim_thompson5910

solving x+2 > 0 for x gives x > -2, which is the domain

- anonymous

Alright, I see :) Now what about the range?

- jim_thompson5910

the left side shoots off to negative infinity (very quickly)
the right side slowly approaches positive infinity. it just keeps growing forever

- anonymous

so the range would be all real #'s or infinity

- jim_thompson5910

all real numbers
I think you're thinking of this notation \(\Large (-\infty,\infty)\)

- anonymous

Yes :)

- anonymous

Thanks!

- jim_thompson5910

you're welcome

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