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)