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Algebra 9 Online
OpenStudy (anonymous):

how do i set this up some help please?

OpenStudy (anonymous):

OpenStudy (.whitedragon.):

is it multiple choice

OpenStudy (anonymous):

Image is way too big :S, can't see much other than a couple of words.

OpenStudy (anonymous):

@Euler271

OpenStudy (anonymous):

@helpme1.2

OpenStudy (anonymous):

@iambatman that'sall thei nfo given to me

OpenStudy (anonymous):

Can you type the problem out?

OpenStudy (.whitedragon.):

@iambatman will u be my fan

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

not really

OpenStudy (anonymous):

and alsom ay i ask how you came upw ith 11.4 i think that'sa nother problemi 'm having?

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

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:\[\log_{1.05}(2) = \frac{ \ln(2) }{ \ln(1.05) }\]

OpenStudy (anonymous):

11.4 is 2 times 5.7

OpenStudy (anonymous):

oh okay understand now

OpenStudy (anonymous):

okay hold on a sec...

OpenStudy (anonymous):

14.21

OpenStudy (anonymous):

yes. so the answer is 2000 + 14.21 years. not sure whether or not you need to round up

OpenStudy (anonymous):

14.21+2000=2014

OpenStudy (anonymous):

looks good

OpenStudy (anonymous):

yay thank you. ih ave one more problem kind oft he same.

OpenStudy (anonymous):

OpenStudy (anonymous):

for this one i have v=1000()^()t not sure what top ut in it

OpenStudy (anonymous):

is it 6%/2 +1? =1.03

OpenStudy (anonymous):

@Kainui

OpenStudy (kainui):

I don't really know what your question is.

OpenStudy (anonymous):

i need help with problem number 2

OpenStudy (kainui):

How can I help you?

OpenStudy (anonymous):

i needh elp solving this problem here...

OpenStudy (kainui):

Where are you stuck?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

@Kainui any idea?

OpenStudy (anonymous):

@ranga

OpenStudy (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.

OpenStudy (anonymous):

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?

OpenStudy (ranga):

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)

OpenStudy (anonymous):

okay so my final answer will belog(v/1000)/2log(1.03)

OpenStudy (ranga):

Yes. \[\Large t = \frac{ \log(V/1000) }{ 2\log(1.03) }\]

OpenStudy (ranga):

If you want you can write numerator as log(V) - log(1000) = log(V) - 3 since log(A/B) = log(A) - log(B)

OpenStudy (anonymous):

okay

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