Solve the simultaneous equations

[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...

 

[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 learned 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
[tex]10x^2+7x+ 1= 10x²+ 20x+ 10[/tex]
so
$$-13x= 9$$

 

[arildno]

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