Can someone tell me what I should do for this problem? I would really appreciate it.
well you could multiply 90,000 by 0.046 then take 221,200 and divide the sum bit. I hope this is right (mathematics isn't my strong point)
by it^
@AZ can you help me?
Sure! Do you know the formula for compound interest?
@AZ i know how to find the rate but other than that i'm lost
\(\Large A = P(1 + \frac{r}{n})^{nt} \) so A = 221,200 P = 90,000 r = 4.6% but first you have to make it as a decimal n = 1 since we're compounding each year t = ?? what we're solving for
can you plug in the numbers and solve for t
P is the principle amount which is what you start off with- the 90,000 the rate or r is 4.6% t is time n is the number of times you compound it A is the final amount including interest
@AZ alright i'm trying it now
@Az I got it. thank you!
Perfect! Let me know if you want me to check your answer :)
@AZ it would be 20 right? 21's to much
That's correct haha but did you go plugging in numbers?
@AZ yes i did
you don't need to tag me each time haha, I get a notification each time you reply but you didn't need to brute force it
so let me show you how to calculate it using the formula
oops sorry. thanks for the help tho <3
you plugged in the numbers I'm sure and got \(\large A = P(1 + \frac{r}{n})^{nt}\) \(\large 221,200 = 90,000(1 + \frac{0.046}{1})^{1t}\) \(\large 221,200 = 90,000(1 +0.046)^{t}\)
yeah that's what i got
\(\large 221,200 = 90,000(1.046)^{t}\) and then we divide 90,000 on both sides \(\large 2.4577 = (1.046)^{t}\)
following along still?
yes i am
so now you have to know that if we take the natural log on both sides, we can bring down that exponent \( ln(a^b) = b ~ln(a)\) so \( \large ln(2.4577) =ln( (1.046)^{t})\) and we get \( \large ln(2.4577) =t~ ln(1.046)\) and now to solve for 't' you just divide ln(1.046) on both sides
does that make sense now?
yes thank you
perfect so basically t = ln(2.4577) / ln(1.046) and ta-da that's all
you're a life saver. i can not thank you enough for your help
of course! it was my pleasure :)
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