How to solve calculations like these in an exam fastly
@AriHoresh
(1.03)^24
or (2.07)^79
or 72^35
Hi,
you need to break down the calculation into easier steps, and estimate the answer
(72)³⁵ = ((9 * 8)⁵)⁷
= (60000 * 33000)⁷
= (200 * 10⁷)⁷
= (2 * 10⁹)⁷
= 128 * 10⁶³
= 1,28 * 10⁶⁵
I’ve estimated a lot of the exponents but i still found a value close to the actual one (1,0 * 10⁶⁵)
1 Like
how did you find 8 to the power of 5
wow
if (a^n)^m = (a^nm), then (72)³⁵ = (72⁵)⁷ = ((9 * 8)⁵)⁷
yes I understand that, but I was wondering how you solved -relatively accurately-
(9*8)^5 to equal (60000 * 33000)