C2 Question ( again )

Solve
log(x+1)-logx=log7

using the log law that log x - log y = log x/y

log x+1/x = log 7

x+1/x = 7

7x = x + 1

6x = 1

x = 1/6

Reply 2

silent ninja

well using the rules of log, the lhs become log((x+1)/x) = log7

so
(x+1)/x = 7

solve from there

Reply 3

hey!
just out of interest, what would you do if the base for one of the terms is different? for example

Solve
log₂(x+1)-log₃x=log₂7? tnx in advance... :smile:

Reply 4

Hello!
to solve
log₂(x+1) - log₃x = log₃7
you could use the change of base rule
log_ab = (log_cb)/(log_ca)
so
log₂(x+1) = log₃(x+1) / log₃2
and so your question is now

log₃(x+1) / log₃2 - log₃x = log₃7

love danniella

Reply 5

danniella

Hello!
to solve
log₂(x+1) - log₃x = log₃7
you could use the change of base rule
log_ab = (log_cb)/(log_ca)
so
log₂(x+1) = log₃(x+1) / log₃2

and so your question is now

log₃(x+1) / log₃2 - log₃x = log₃7

love danniella

Danniella it was log₂ 7

So the question looks like

log₃(x+1) / log₃2 - log₃x = log₃7 / log₃2

log3(x+1) / log32 - log3x = log37 / log32

so

log3(x+1) / log32 - log37 / log32 = log3 x

[log3 (x+1) - log3 7]/ log3 2 = log3 x

log 3 (x+1)/7 / log3 2 = log3 x

log3 (x+1)/7log3 2 = log3 x

log3 (x+1)/ log3 2^7 = log3 x

log3(x+1)/log3 x = log3 128

Thats as far as i can get.

Reply 6

oh thats too bad! I totally forgot about that rule! Must revise i guess... tnx danniella! :biggrin:

Reply 7

Um.
Actually, neither of us was right. The "odd log out" bit was the
log₃x, which would change to
log₂x / log₂ 3.

Oops.
So the actual question is now
log₂ (x+1) - (log₂x)/(log₂3) = log₂7.

We'll get there eventually!

love danniella

Reply 8

Well i tried the base 2 and didnt get very far so i went for the base 3.

Reply 9

insparato

Well i tried the base 2 and didnt get very far so i went for the base 3.

er, say what?
:confused:

love danniella

Reply 10

danniella

er, say what?
:confused:

love danniella

I think that he ment that he tried to change everything to 2 and having failed to do so, he changed everything to base 3 and worked it out...

Reply 11

Um...

I think at this point I'm glad you two seem to be able to understand each other. Obviously it's just me... :p:

love danniella

Reply 12

Sorry Danniella i can be abit on the non explanatory side when it comes to things that ive used too many times.

Yes well the obvious first step is to change the bases of the logs so that all of the logs are in the same base. However when attempting to change what needed to be changed for everything to be in base 2 i didnt get very far. So i tried changing what needed to be changed into base 3 to see if that helped. I got alil further.

Reply 13

OK here is a very stright foward question which just poped up in my head...
what exactly is point of inflection? i know it is the point where f'(x) = 0, f''(x) =0 and f'''(x) is not equals to 0 but what exactly is it? it cant be a point where it has repeated root because y = x² doesnt have a point of inflection... I never understood what it ment all this time :frown:

Reply 14

Hello!

http://en.wikipedia.org/wiki/Stationary_point

is probably your best bet.

love danniella

Reply 15

OK... this is what i came across few minutes after i posted the question... according to some site i found via google...
"point of inflection is the point where the gradiet of the curve is 0"... basically, the curve is parrel to the x-axis at that point... it still is a bit doggy :frown:

but i guess that will do for now... :biggrin:

Reply 16