Graphing Logs! (2 Viewers)

[SoCal]

I have been trying to graph the equation y=log(x-5) and I have got half way there.

Domain: x-5>0
x>5
so asymptote at x=5

Now using x-5=e^y
when y=0
x-5=e^0
x-5=1
x=6
so cuts x-axis at (6,0)

Now I need to plot a point of the graph. How can I do that? Thank you.

 

[evilc]

sub x = e + 5 into the equation: y = log(x-5), and you get:
y = log(e + 5 - 5) = log(e) =1
so, you have the point (e + 5, 1)

 

[...]

hmm...u got the x-axis.good...

umm...u can get a -ve y value...and in terms of that "logly", it will be decreases at a very rapid rate...so u would have the graph coming from -ve y--> +ve y in a very very short space... and after it hits (6,0), the graph will be slowly increasing...

the pic is below...and it should read y=log(x-5) instead of 1

 

[mitochondria]

do you know how to graph a y=log x graph? and y=log9x-50 graph is just a y=logx graph moved 5 units to the right.. or otherwise, you can graph y=logx first and move the y-axis 5 units to the right.. hops that help..

if you want to plot a point.. you would probably want to do (5+e, 1) but there is probably no point of doing this (excuse the pun )

aww.. 2 replies in the time i was typing my message!?

 

[SoCal]

Thank you all for your fast replies.

Evilc, how did you know to substitute x in as e + 5? Also in this line: y = log(e + 5 - 5) = log(e) =1, how do you know log(e)=1? I get an error on the calculator.

Thank you for the graph "...", that is how I thought it would look.

Mitochondria, that is why I have an asymptote at x=5, thanks for going over it again anyway. Also, I have always been taught to supply a point on the curve, is that not advised or something?

Thanks again.

 

[...]

log(e) doesn't equal to one...ln(e) =1

log is base 10 remember

u got error cos ur suppose to type in e1 instead just e

 

[evilc]

yeah, i used "log' to mean 'log base e' instead of "ln" in this context, i was going to write a note about it in my post, but i didnt think it was necessary.

 

[evilc]

Merethrond, remember how: y= Ln(e) can be rewritten to:

e^y = e ...which has the obvious solution of 1, so therefore Ln(e) is equal to 1.

now, since the function is: y = Ln( x - 5), you need to sub in a value that will make the part of the function inside the brackets equal to 1, ie. e + 5

My solution applied to Log base e, not log base 10. I assumed (wrongly) that log was log base e, probably coz some textbooks i have used just write log instead of Ln when dealing with log base e.

From now on, I'll use log(x) as meaning log base 10 of x.


So, to find a point on the graph of log( x - 5), sub in a value of x so that the stuff inside the brackets is equal to the base of the log (10), so sub in x =15, as 15 - 5 = 10, and so the point on the graph is in fact (15,1) which you would expect, by looking at ...'s graph.

 

[...]

is this ln or log??

cos u can get 2 slightly different graph...

 

[SoCal]

Originally posted by ...
log(e) doesn't equal to one...ln(e) =1

log is base 10 remember

u got error cos ur suppose to type in e1 instead just e

Sorry dude, that is supposed to say log(base e) or ln, not log to the base 10.

 

[evilc]

oh right, so the original graph was to base e? if so, disregard the second part of my last post

 

[...]

well, that graph i had there was very similiar to ln(e) anyway...

just get a lttle rigid...

 

[SoCal]

Evilc, I was forgetting that ln x e = 1, so that is why I couldn't understand what was going on. Thank you again for your explanation. It is all good now.

Sorry for typing down the wrong thing "...". I was just a bit lazy.

 

[SoCal]

I have another question. If you have the equation:

y=lnx

and you convert it to

x=e^y

is that just a rule or can you derive it somehow?

 

[...]

its a rule...

 

[SoCal]

Originally posted by ...
its a rule...

Thanks.

 

[evilc]

yeah, by defenition, a logarithm is the inverse of an exponent, eg.

10^2 = 100,
so Log 100 = 2 which in words means "the power that 10 is raised to to give 100 is equal to 2."

So, taking this principle and applying it to y = Lnx, "the power that e is raised to give x is equal to y."
we can see that: e^y = x and y = Lnx are equivalent

I hope that made some sense

 

[mitochondria]

you can derive it
y = lnx

--> by raising both sides as powers of e, you have

e^y = e^(lnx)

but essentially expotential and logarithum are inverse of each other, therefore by raising lnx as a power of e, they "cancle" each other out, and you have:

e^y=lnx

and this explains why when you substitude x=e into y=lnx, you'll get y=1 because:

y=ln(x)

for x=e

y=ln(e)

e = e^1

therefore:

y=ln(e^1)

and as i said, they sort of "cancle" out each other, so you have y=1 I hope that helps


Also.. just a note that might be of interest, mathematically if log is written on its own, that is, the base is not recognised, it is treated as ln...

Good luck and all the best with ya HSC



*how come whenever i'm posting someone is posting + with similar content?* hehee..

 

[evilc]

haha, sorry mitochondria..thats twice i've done it to you.. your explanation is much better than mine.

 

[mitochondria]

no no no, i like yours evilc i was think about using your way to do it because it's easier to understand... hehee.. i wasn't sure if the "canceling" idea is simple enough for everyone too.. still still... it's okay, lol, is civil engineering fun? meee like people like you who don't mind helping us HSC studnets *a big round of applause for evilc*