But I can't remember my maths lessons. How do I rearrange the below formula to make x the subject:
Code:y = 2x
Printable View
But I can't remember my maths lessons. How do I rearrange the below formula to make x the subject:
Code:y = 2x
This is what it came into my mind:
Might be wrong. :DCode:taking log() on both sides
log(y) = log(2^x)
log(y) = x.log(2)
ie., log(y)/log(2) = x
ie. log(2.y) = x
jemidiah, Milk, etc.. are the pioneers in math. :thumb:
:wave:
This looks like a fun tool, and it's free: http://www.microsoft.com/downloads/e...f-9e242b794c3a
This isn't a snide comment about how you should use that to find the answer, I genuinely think it's a nice tool. I think log is the correct answer.
log(y)/log(2) = x works :D
The question is, why does this work? What is the proof?
"It just does - deal with it." - Sir Isaac Newton.
It works because a logarithmic function is an inverse function of the exponential function.Quote:
Originally Posted by visualAd
It's like how division is inverse of multiplication and saying that if y=5*x => x = y / 5
It works because one of the most basic rules of logarithms is that
log(px) = p log(x)
This makes sense if you see that the logarithm of x is just the number to which the base (usually 10) must be raised to produce x. For example, the logarithm of 1000 is 3, because the base (10) has to be raised 3 times (10^3) to produce 1000. The logarithm of 10^3 (= 3) is therefore the same as 3 times the logarithm of 10. [ In other words: log(10^3) = 3 log(10) ]
So just take the logarithm of both sides and rearrange.
Ah logarithms!! Along with derivatives and integration, these spoilt my math classes in college...
.
Excellent, thanks for the maths lesson. It kind of rings a bell of a A Level lesson sometime in 1999 :D
1999? ... whippersnapper
Well here is my test. I took the formula for calculation of compound interest:
Based on the above, I was able to rearrange the formula to make n the subject:Code:M = P (1 + i)n
- M = maturity amount
- P = initial amount
- n = number of intervals (e.g. years)
Can this be simplified any more?Code:n = log(M / P) / log((1 + i))
:wave:Code:n = log((M / P) * (1 + i))
Is this another rule:
log(a) / log(b) = log(ab)
I think, I'm wrong. : http://www.andrews.edu/~calkins/math...xts/numb17.htmQuote:
Originally Posted by visualAd
I'll double check it. :thumb:
No, I'm pretty sure it's log(a) / log(b) = log(a - b).
Sorry for the mistake. I have mixed up everything in my mind :bigyello:Quote:
Originally Posted by NickThissen
Right ?Code:log(a) + log(b) = log(a*b)
:wave:
Yeah that was it. Mine was wrong too, I think it should have been log(a) - log(b) = log(a/b)
I think, the correct answer is:
Am I right ?Code:log(a)/log(b) = log a to the base b
:wave:
Yes, it's a change of base, but that doesn't really simplify the expression. In fact I would say that
log1+i(M / P)
is less simple then
log(M/P) / log(1 + i)
But that's just me I guess :)
All your base...
This belongs in the maths forum. I will request that it be moved there.
Neatfreak...
The problem is already solved and Shaggy already started to derail this thread.
Have to agree with moonman239, there is some useful information in this thread that may be of use to other people, so going to move it over to the Maths Forum.
Gary
Logarithm formulas have a bunch of symmetries with exponential formulas, since they're inverse functions. Examples:
y = e^x
<=>
x = ln(y)
e^(a+b) = e^a * e^b
<=>
ln(a) + ln(b) = ln(a*b)
e^(ax+by) = (e^x)^a * (e^y)^b
<=>
ln(x^a * y^b) = a*ln(x) + b*ln(y)
e^(x ln(b)) = b^x
<=>
ln(x)/ln(b) = log_b(x)
Of course there are others.