# 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