Question

Approximating 10^(-0.83) =?

Answer 1

o

Answer 2

no :/

Answer 3

almost

Answer 4

\[\LARGE 10^{-6.83}\]
I have to find for this...

Answer 5

\[\LARGE \frac{10^{-6}}{10^{0.83}}\]
Did it this way

Answer 6

it'll be slightly greater than 0.1

Answer 7

10^(-0.83) i.e.

Answer 8

sab Q-tiyapa hai :)

Answer 9

You could use series expansions, but you would have to calculate (-0.83)^n, so if n is even, there is a complex number.
\[10^{x}=\sum_{n=0}^{\infty} \frac{x^{n}\ln^{n}(10)}{n!}\]
There is the linear approximation:
\[L(x)=f(a)+f'(a)(x-a)\] the derivative of 10^x is ln(x)*10^x, so for negative numbers you would end in complex numbers again.
These are the only ones I know.
(Sorry for my english)

Answer 10

complicated question.