how do i set this up some help please?
is it multiple choice
Image is way too big :S, can't see much other than a couple of words.
@Euler271
@helpme1.2
@iambatman that'sall thei nfo given to me
Can you type the problem out?
@iambatman will u be my fan
you want to find t when D = 11.4 trillion. twice the initial debt in 2000. \[11.4 = 5.7(1.05^t)\]\[2 = 1.05^t\]\[the\ inverse\ of\ y = a^x\ is\ \log_a(y) = x\]we are solving for x (which is t in our case) \[\log_{1.05}(2) = t\]do you know how to use a funky log base with your calculator?
not really
and alsom ay i ask how you came upw ith 11.4 i think that'sa nother problemi 'm having?
sorry for the bad keyboard grammar. i was wanting to know how you came up with 11.4 that is another problem i have is how to find that number before getting started?
this is how it's done. the log button on your calculator is log base 10. to use base "a" instead, we would do this: \[\log_a(x) = \frac{ \log_{10}(x) }{ \log_{10}(a) }\] this also works with ln button (log base e). as long as you use the same log base in the numerator and denominator. so:\[\log_{1.05}(2) = \frac{ \ln(2) }{ \ln(1.05) }\]
11.4 is 2 times 5.7
oh okay understand now
okay hold on a sec...
14.21
yes. so the answer is 2000 + 14.21 years. not sure whether or not you need to round up
14.21+2000=2014
looks good
yay thank you. ih ave one more problem kind oft he same.
for this one i have v=1000()^()t not sure what top ut in it
is it 6%/2 +1? =1.03
@Kainui
I don't really know what your question is.
i need help with problem number 2
How can I help you?
i needh elp solving this problem here...
Where are you stuck?
well so far i have v=1000()^()t where i have those parenthesis is wherei 'm having problems not sure what answer goes in thosetw o places
@Kainui any idea?
@ranga
Just plug in all the given numbers. The following numbers are given in the problem: P, r, n. Plug the numbers into the formula and you will have V as a function of t.
okay well i have v=1000(0\)^6()t so to get the numberi n the parenthesis do i do 6%/2+! to get 1.03?
P = 1000 r = 6% = 0.06 (r should be in decimal in this formula) n = 2 (compounded semi-annually means compounded TWICE a year and so n = 2)
okay so my final answer will belog(v/1000)/2log(1.03)
Yes. \[\Large t = \frac{ \log(V/1000) }{ 2\log(1.03) }\]
If you want you can write numerator as log(V) - log(1000) = log(V) - 3 since log(A/B) = log(A) - log(B)
okay
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