Logarithms

[footprints]

Sovle the simultaneous equations
log_2 (x-14y) = 3
lgx - lg(y+1) = 1
How do i start?

[misogynisticfeminist]

for the first equation, make it

2^3= x-14y

then carry on from there......

:smile:

[HallsofIvy]

Of course, without knowing what "lg" means we can't help you with the second equation!

[footprints]

I suppose lg means log_{10} ? I got the question from my book.

[Hurkyl]

Weird. I'm most familiar with lg denoting log base two, from my computer science courses.

[footprints]

Well u might be correct considering that u are probably smarter than me. But i was thaught that lg is log_{10} if the base isn't stated.

[arildno]

lg is used, for example in certain fluid mechanics formulae, as the Briggsian logarithm, that is, log_{10}

[footprints]

The answer for the question is x=15, y=1/2

[arildno]

Since you've learnt to use "lg" as log10, stick with that!
The second equation can then be rewritten as:
\frac{x}{y+1}=10
Do you agree with that reasoning?

[footprints]

Yes. Thank you

[footprints]

How about this one. Solve 1 + 2 lg (x+1) = lg (2x+1) + lg (5x+8)
Sorry but i still haven't got the hang of log yet

[arildno]

Look first at your RIGHT-HAND side:
Can you write the sum of two logs as a single log?

[footprints]

ok so i got lg(10x^2 + 21x + 8)
Then i will get 10+(x+1)^2 = (10x^2 + 21x + 8)right?

[arildno]

Right, so you can use that expression as your right-hand side instead (agreed?).
Now, consider the 2lg(x+1)-term on your original left-hand side.
Can you rewrite that into log(something..)

[arildno]

No, your suggestion at exponentiating the equation is wrong, even though you made a correct rewriting of your right-hand side

[footprints]

Isn't the left hand side in a log form already? Except the 1.

[arildno]

It is completely wrong:
We have:
1+lg((x+1)^{2})=lg(....)
We must move the log term on the left-hand side over and get:
1=lg(\frac{(...)}{(x+1)^{2}})
Or :
10=\frac{(...)}{(x+1)^{2}}

Do you see the difference?

I've used (...) to denote what stood on the right-hand side.

[footprints]

Ahhh... Finally i get it. I can solve it from here.

[MiGUi]

What notation you use for logarythms in base "e" and in base "10"? In Spain we use "ln" for the first, and "log" for second. I think that your notation is not the same...

[footprints]

Same for me in my country

[HallsofIvy]

I would do 1 + 2 lg (x+1) = lg (2x+1) + lg (5x+8)
by rewriting it as lg(2x+1)+ lg(5x+8)- 2log(x+1)= 1 so
lg((2x+1)(5x+1)/(x+1)2)= 1 which is the same as

\frac{(2x+1)(5x+1)/(x+1)^2}= 10
or
(2x+1)(5x+1)= 10(x+1)^2
which is
10x^2+7x+ 1= 10x2+ 20x+ 10
so
-13x= 9

[arildno]

Halls: It is (5x+8) rather than (5x+1)..