1. ## Simultaneous equation help

I have to determine the values of a and b in this equation. I'm told to use simultaneous equations. But I can't find my notes on them, although I'm not sure that would help me anyways. The log bit also adds to the confusion

The equation is:

log M = av + b

Now I know M, it is 45,000 for substance O, and is 17,200 for substance Y.

I also know v, it is 0.000258 for substance O, and is 0.000916 for substance Y.

If you could use the substitution method, that would be great as that is the one I found easier in class. Thank you. Please say if you need any more information.

2. Originally Posted by projo
I have to determine the values of a and b in this equation. I'm told to use simultaneous equations. But I can't find my notes on them, although I'm not sure that would help me anyways. The log bit also adds to the confusion

The equation is:

log M = av + b

Now I know M, it is 45,000 for substance O, and is 17,200 for substance Y.

I also know v, it is 0.000258 for substance O, and is 0.000916 for substance Y.

If you could use the substitution method, that would be great as that is the one I found easier in class. Thank you. Please say if you need any more information.
if that equation satisfy both substances O and Y, then simultaneous equation is

$\begin{array}{lcl} \log 45000 & = & 0.000258 \, a + b \\ \log 17200 & = & 0.000916 \, a + b \end{array}$

now, can you solve the system?

3. Originally Posted by kalagota
if that equation satisfy both substances O and Y, then simultaneous equation is

$\begin{array}{lcl} \log 45000 & = & 0.000258 \, a + b \\ \log 17200 & = & 0.000916 \, a + b \end{array}$

now, can you solve the system?

No, not really. I didn't even get as far as that.

4. Originally Posted by projo
No, not really. I didn't even get as far as that.

well, i just used the info you gave about O and Y and substitute it to the equation..

so, that system can be solved easily.. you can do your substitution.. but the easiest ways is to subtract one from the other, say equation 2 from equation 1.. in this manner, the b's cancels out and only a remains, and then solve for a. after solving for a, substitute it to any of the equation and solve for b.

5. What, so I get should get

log 27800/-0.000658 = a?

Just like to say, Maths is not my thing, as you may have guessed, so forgive me if I am crap. I do chemistry instead.

6. no..

take note of these things about logarithms:

$\log a \pm \log b \neq \log (a \pm b)$

in fact, this is the correct one:

$\log a + \log b = \log (ab)$ and $\log a - \log b = \log \frac{a}{b}$

where $a,b > 0$

also take note of this: $\log a^x = x \cdot \log a$ and do not confuse it with $(\log a)^x$.. they are different things.. the first one is, $x$ is raised on $a$, while the second one, $x$ is raised on the whole logarithm..

thus, from the system, you should get

(on the left hand side)
$\log \frac{45000}{17200} = \log \frac{225}{86}$

and on the right hand side, just the usual subtraction..

7. Thanks for the info on logs.

so a=0.000258-0.000916(log45000/log17200)

8. Originally Posted by projo
Thanks for the info on logs.

so a=0.000258-0.000916(log45000/log17200)
no, it should be like this.. $a = \frac{\log 45000 - \log 17200}{0.000258 - 0.000916} = \frac{\log \frac{45000}{17200}}{-.000658}$

now, all you have to do is to simplify it (you can also use your calculator)..

you should get $a = - 634.7782$

now, plug that in to any of the equation to solve for b.. Ü

9. Oh my God. Thank you loads mate. That has really really helped. Now I can finally lay this silly maths/chemistry assignment to sleep.

10. Originally Posted by projo
Oh my God. Thank you loads mate. That has really really helped. Now I can finally lay this silly maths/chemistry assignment to sleep.
no problem.. Ü