Solve the simultaneous equations

1. Oct 23, 2004

footprints

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

2. Oct 23, 2004

misogynisticfeminist

for the first equation, make it

$$2^3= x-14y$$

then carry on from there......

3. Oct 23, 2004

HallsofIvy

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

4. Oct 23, 2004

footprints

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

Last edited: Oct 23, 2004
5. Oct 23, 2004

Hurkyl

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

6. Oct 23, 2004

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.

7. Oct 23, 2004

arildno

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

8. Oct 23, 2004

footprints

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

9. Oct 23, 2004

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?

10. Oct 23, 2004

footprints

Yes. Thank you

11. Oct 23, 2004

footprints

Sorry but i still haven't got the hang of log yet

12. Oct 23, 2004

arildno

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

13. Oct 23, 2004

footprints

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

14. Oct 23, 2004

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..)

15. Oct 23, 2004

arildno

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

Last edited: Oct 23, 2004
16. Oct 23, 2004

footprints

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

17. Oct 23, 2004

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.

18. Oct 23, 2004

footprints

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

Last edited: Oct 23, 2004
19. Oct 23, 2004

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...

20. Oct 23, 2004

footprints

Same for me in my country