precalc-natural logs?
t=1/c[ln A - ln(A-N)]
t=time in weeks it takes to achieve mastery of a portion of a task, A is the maximum learning possible, N is the portion of the learning to be achieved, and c is a constant used to measure an individual's learning style.
1. express the formula so that the expression brackets is written as a single logarithm.
so first i did t = 1/c [ln(A/A-N)]
then t = ln [A/(A-N)] ^(1/c)
does this work? cause when there is subtraction it condenses to division. then when a number is in front of ln, you can but it as an exponent?
part 2:
how many weeks will it take a chimp to master 30 signs when 65 signs is the max and c=.03?
EDIT
23 minutes ago
using the formula that i condensed, it doesn't work with part 2. i get like .000000114 weeks....
22 minutes ago
so i did something wrong.. cuz i'm pretty sure .000000114 weeks is incorrect, which is what i got when substituting the info into the formula i condensed.
Comments
Your formula for #1 looks right:
t = 1 / c [ln A - ln (A - N)] = 1 / c ln [A / (A - N)] = ln [A / (A - N)]^(1/c)
so far so good...
when I plug in A = 65, N = 30, and c = .03:
t = ln [65 / 35]^(1/.03) = ln [1.857^33.33] = 20.63 weeks
did you take the log before you raised it to the power? that could be your mistake
t = (1/c) (ln(A) - ln(A - N))
t = (1/c) ln(A/(A - N))
t = ln((A/(A - N))^(1/c))
your answer is correct
remember the index laws
when you multiply indices with the same base, you add the exponents
when you divide indices, you subtract the exponents
a^b * a^c = a^(b + c)
a^b / a^c = a^(b - c)
logarithms are the opposite,
when you add logarithms with the same base, you multiply the numbers inside the logarithm (forgotten what there called)
when you subtract logarithms with the same base, you divide the numbers inside the logarithm
loga(b) + loga(c) = loga(bc)
loga(b) - loga(c) = loga(b/c)
the power rule is just another logarithm law
bloga(c) = loga(c^b)
t = lnx
x = (A/(A - N))^(1/c)
x = (65/(65 - 30))^(1/c)
= 1.857...^(1/0.03)
x = 915188195.5..
t = lnx
t = 20.63 weeks (2dp)
what you did was take the whole logarithm to the power of 1/c
only whats inside the logarithm is raised to this power
a million) 3000 = 2(2 + e^2x) 1500 = 2 + e^2x 1498 = e^2x Take organic log of the two aspects ln is organic log. ln(1498) = ln(e^2x) ln(1498) = 2x x = ln(1498)/2 2) 10^(x-6) = 7/5 x - 6 = (7/5)^(a million/10) x = (7/5)^(a million/10) + 6 3) log(15xy) - log(3z^2) 5*6*.5*log(y/2xz) = 15log(y/2xz)