# Easy math problem

Registered User regular
edited September 2008
Ok, so I haven't taken a math course in awhile and I have some chem homework due next week and I just wanted to double check my answer with you guys. I've gone over it a bunch of times, but I still feel like Im making a stupid mistake.

Y = (Z/X) + 2 Solve for X

I came up with YZ - 2Z = X

Also, there's a problem asking for the antilog of -3.8456, how do I do this on my TI-83-plus?

TheMorningStar on

## Posts

• Registered User regular
edited August 2008
I think you are making a mistake. After you subtract 2 from both sides you need to multiply both sides by X, then divide by Y-2 to get:

X=Z/(Y-2)

Boutros on
• Registered User regular
edited August 2008
I think you missed up, just a tad bit for coming up with; X = YZ-2Z.

Y = (Z/X) + 2. ---> original

Y - 2 = (Z/X).
> subtract 2 from both sides

X = (Z/Y - 2). ----> switching positions of the (X) and (Y-2)

I think what you did wrong was at the part where Y - 2 = (Z/X).

Xano on
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• Registered User regular
edited August 2008
also for antilogs, I don't think there is a button on the TI-83 to do that.

I'm pretty certain you just do 10 ^ -3.8456.

I'd rather have someone else confirm it though.

Xano on
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• Registered User regular
edited August 2008
Thanks for the replies, I understand now! But now I've ran into one more problem I cant figure out:

How many liters of a 10% alcohol solution must be mixed with 30 liters of a 50% solution to get a 20% solution?

I remember doing a problem almost exactly like this two semesters ago, but now I have no idea where to even start. Help!

I figured it out! Now I just need to do more investigating into the antilog thing...

TheMorningStar on
• Registered User regular
edited September 2008
The antilog of -3.8456 depends on the base. Usually people use the natural base e or base 10. In general though, the antilog can be defined as b^x, where b is the base and x is -3.8456 in your case.

Aero on
• Registered User regular
edited September 2008
Thanks for the replies, I understand now! But now I've ran into one more problem I cant figure out:

How many liters of a 10% alcohol solution must be mixed with 30 liters of a 50% solution to get a 20% solution?

I remember doing a problem almost exactly like this two semesters ago, but now I have no idea where to even start. Help!

I figured it out! Now I just need to do more investigating into the antilog thing...

For the liters problem, you just need to think that .1x + .5y = .2z where x + y = z, where we know y is 30, so it becomes a system of equations problem with two variables and two equations.

0.1x + 15 = 0.2z
1.0x + 30 = 1.0z

There's a couple of ways to solve this, but I feel kind of weird doing someone else's homework, so I'll let someone else finish up, or tell me I'm doing it totally wrong.

Malyonsus on
• What's on sale? Pliers!Registered User regular
edited September 2008
Xano wrote: »
I think you missed up, just a tad bit for coming up with; X = YZ-2Z.

Y = (Z/X) + 2. ---> original

Y - 2 = (Z/X).
> subtract 2 from both sides

X = (Z/Y - 2). ----> switching positions of the (X) and (Y-2)

I think what you did wrong was at the part where Y - 2 = (Z/X).

The bold part is actually multiplying both sides with X and dividing both sides by (Y-2). You can't do that if Y-2 = 0. So you have to do a special case for Y=2 and exclude it explicitly before doing this operation.

(The special case goes: if Y=2, then 0 = Z/X, so Z=0 and X can be anything.)

Grobian on