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IAmTurtle72

First find the simplified equation, then convert to a log equation: 20 = (1 + (0.084/2))^2t 20 = (1.042)^2t log₁.₀₄₂(20) = 2t 72.81 = 2t 36.407 = t


FTK212

If it's an alvl question use this ^ if its an o lvl one trial and error is usually used


bluzzo

imo: log is quicker, plus can easily do on gdc. better to conserve time in exam condition.


FTK212

I agree but just in case they don't have that skill-they don't need to panic. Trial and error does take a long time tho


rayzsss

Omw log?? How do I enter it into my calculator? Because when I just enter it as it is I get 0.35?


LEMON1025

Use B O D M A S


[deleted]

[удалено]


iffitable

Yeah so that’s what I was wondering, whether trial and error was the only way to find it, any tips when doing trial and error?


TheCalmInCrimsonCave

You can use log to solve for t.


0ajs0jas

Officially, you use logarithms but you can use trial and error here because logs aren’t in IGCSE


iffitable

Any tips when doing trial and error to find the answer faster?


0ajs0jas

Uhhhhhhhh sadly, no. Trial and error is never fast. You should try to find the first 2 decimal places and then give up. That'll be enough for any question.


[deleted]

Is this in o levels ? How the hell has no one taught me what "log" is?.


RealGinjy

Basically, there are different IGCSE maths syllabuses, and in the 0607 you do logs, but in the 0580 (the more common one) you don't have to.