Question

antilog -0.699
can someone plz help me in verification​

Answer 1

I'm Considering the base of antilog as 10

$antilog_{10}( - 0.699) = {10}^{ - 0.699} \approx {10}^{ - 0.7}$

Let,

$n = {10}^{ - 0.5 - 0.2}$

$n = \sqrt{ \frac{1}{10} } \sqrt[5]{ \frac{1}{10} }$

$n = \frac{ \sqrt[5]{10000} \sqrt{10} }{100}$

$n \approx6.31 \times \frac{3.16}{100}$

$n \approx0.199$

Thus, we can conclude that:

$\boxed{ antilog( - 0.699) \approx0.199 \approx0.2 }$