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Simple algebra problem

Brian-M

Daydreamer
Brian-M
Joined
Jul 22, 2008
Messages
8,044
Okay, this is probably going to seem blindingly obvious once somebody points out how to do it, but for some reason I can't seem to figure it out.

If you have an equation such as YX = Z, how do you solve for X?

For example, with 2x = 3, X has to be 1.5849625, but I can't figure out the proper way to get that answer. (I got it by trial and error to find each successive digit.)

So, anybody know the right way to get the answer?
 
You're going to want to use logarithms... and the change of base formula ... and a calculator.

Cheers.

Edit: I'm sure someone will be along shortly to actually solve it. I'm posting from my phone and it's the middle of the night, but I figured I could at least get you going down the right path. Once you have it solved for x you'll likely need a calculator or log table to find the numerical value. These have a base of 10 or e, which is why you'll need to change the base. Cheers again.
 
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2^x = 3
x = log(2) 3
x = log(e) 3 / log(e) 2 (The 2 on the bottom comes from the original base)
x = 1.0986123 / 0.69314718
x = 1.5849625

Or if you only have a book of common logarithms, you convert to base10 instead:

x = log(10) 3 / log(10) 2
x = 0.47712125 / 0.30103
x = 1.5849625
 
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BowlOfRed said:
x = log(2) 3
x = log(e) 3 / log(e) 2 (The 2 on the bottom comes from the original base)
Click to expand...

Thanks for that. I've never done much with logorithms, so it wasn't obvious to me. But still, I can see how the log(2) makes sense, even though my calculator only does log(10) and log(e).

I'm not sure why the log(e) solution works, but it does. (I guess I'll probably have to look up natural logarithms on a math site and learn more about it.)

ETA: Looks like GodMark2's link explains that pretty simply.
 
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Just remember the definition.

The logarithm of any number, x, is the power to which the base must be raised to get x
 
Brian-M said:
Thanks for that. I've never done much with logorithms, so it wasn't obvious to me. But still, I can see how the log(2) makes sense, even though my calculator only does log(10) and log(e).
Click to expand...

Exactly. Because we don't usually have access to logs of arbitrary bases (like 2 in this case), we have to convert to a base we do know. "e" and "10" are the two bases you'll be able to find on a calculator.

There's no difference between choosing between e or 10, it's just to get something that you can calculate.
 
Some fairly cheap calculators, such as the Casio fx-83GT PLUS, now have three log buttons, one marked 'log' (base 10) one marked 'ln' (base e) and one marked 'log .| ' (the .| are actually a small filled in square and a larger hollow rectangle).

When you press that third log key, you first enter the base and then the number you want the log of.

It's most useful for log base 2 calculations (often written lg ) where you want to know how many bits are required to hold a certain amount of information. How many bits do I need to hold 1000 different values?

lg(1000) is about 9.966 so the answer is 10 which we know is correct as a 10-bit number can indeed express 1024 different values.
 
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ceptimus said:
It's most useful for log base 2 calculations (often written lg ) where you want to know how many bits are required to hold a certain amount of information.
Click to expand...

That's pretty much the kind of thing I was thinking of when I discovered I didn't know how to work out what the exponent should be.
 
Corsair 115 said:
All right, the answer is, "The gods are angry." Wait, what was the question again?
Click to expand...

If the gods are angry with you, then the original question just became irrelevant. You need to head straight for the nearest lightning-proof bunker. (Or lightning-proof boat, in case Poseidon or Neptune get involved.)
 
Brian-M said:
Okay, this is probably going to seem blindingly obvious once somebody points out how to do it, but for some reason I can't seem to figure it out.

If you have an equation such as YX = Z, how do you solve for X?

For example, with 2x = 3, X has to be 1.5849625, but I can't figure out the proper way to get that answer. (I got it by trial and error to find each successive digit.)

So, anybody know the right way to get the answer?
Click to expand...

I haven't seen a general solution so ehre it is:


y^x=exp (x*ln (y)) so solving for Z:

exp (x*ln (y))=z

Assuming z<=0 there is no solutions (at least in the real plan, in the complex plan it is a different story)
For z>0 then you can take the ln on both side : since ln(exp(k))=k then
you finally get x*ln (y)=ln(z) and so x=ln (z)/ln (y)

taking z=3 and y=2 you get x=ln(3)/ln (2) and so the same solution as everybody did.

I thought the general solution would be important to state , because when solving it is incomplete if you do not mention this is solvable only for z>0. At least when i took maths we only got half point if we forgot to mention that.
 
GodMark2 said:

http://latex.codecogs.com/gif.latex?\begin{align*} 2^x&=3\\ \ln2^x&=\ln3&&(1)\textsl{Log both sides}\\ x\ln2 &=\ln3&&(2)\textsl{Extract exponent from logarithm}\\ \frac{x\ln2}{\ln2}x&=\frac{\ln3}{\ln2}&&(3)\textsl{Divide by constant} \\ x&=\frac{\ln3}{\ln2}&&(4)\textsl{Cancel fraction} \end{align*}

...Don't know why I can't get the image to show...
Click to expand...
There's an "x" too many on the LHS of equation (3). For the rest, yes, that's a definite explanation of all log conversions, nothing to add.


Brian-M said:
If the gods are angry with you, then the original question just became irrelevant. You need to head straight for the nearest lightning-proof bunker. (Or lightning-proof boat, in case Poseidon or Neptune get involved.)
Click to expand...
Boat? Not good if Poseidon is angry with you. Just ask Odysseus. ;)
 
ddt said:
There's an "x" too many on the LHS of equation (3). For the rest, yes, that's a definite explanation of all log conversions, nothing to add.
Click to expand...

I was having trouble getting the image to show (the site specifically allows hotlinking, and I've used it before; it just wasn't working: I suspect some type of timeout is involved). I was so frustrated, I forgot to double-check my equations. :blush:

The key:
Logs turn multiplication into addition
log(a*b) = log(a) + log(b)

Logs turn repeated multiplication into repeated addition.
Repeated multiplication is exponentiation.
Repeated addition is multiplication.
Logs turn exponentiation into multiplication.
 
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