How do I solve for Y? X = D * (R**Y) when of course R**Y = R to the power of Y. I can only get as far as X/D = R**Y but now how do I solve for Y? What does Y equal?
2006-11-10 09:47:36 · 7 answers · asked by Tom Heston 2 in Science & Mathematics ➔ Mathematics
You're getting into logarithm. If X/D = R^Y, then Y = log(X/D), where log is the logarithm in base R.
2006-11-10 09:53:17 · answer #1 · answered by Anonymous · 0⤊ 0⤋
One rule of exponents is that as quickly as exponents are raised to a distinctive exponent, they multiply. So if I had (x^2)^2, it could be equivalent to x^4. we want the exponent on n to be a million, so we strengthen the two facet of the equation to the three/2 potential, by way of fact 3/2 situations 2/3 is one. So: n^2/3 = 9 (n^2/3)^3/2 =(9)^3/2 n = 27 stable success at something of your homework!
2016-12-14 05:00:45 · answer #2 · answered by karsten 4 · 0⤊ 0⤋
Most math is way beyond me, but this is the way we solved for roots in BASIC.
a=x^(1/n)
a=answer
x=number
n=root
2006-11-10 10:03:11 · answer #3 · answered by quaver 4 · 0⤊ 0⤋
suppose D > 0, X > 0, R > 0
then R ^ Y = exp(y log(R))
then taking the logarithn leads to:
log(x) - log(D) = Y * log(R)
then if R not equals to 1 :
Y = 1 / R * (log(x) - log(D))
2006-11-10 09:53:52 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Take the logarithm of both sides (natural in this case)...so we have log(X/D) = Y*log(R) so Y = log(X/D) / log(R)
Steve
2006-11-10 09:50:52 · answer #5 · answered by Anonymous · 0⤊ 0⤋
do you mean
x = d * r^y
x/d = r^y
ln [x/d] = y ln r
y = [ ln (x/d) ] / ln r
2006-11-10 09:50:31 · answer #6 · answered by bourqueno77 4 · 0⤊ 0⤋
who cares..you dont use that in real life
2006-11-10 09:55:04 · answer #7 · answered by luckychucky 2 · 0⤊ 3⤋