Question

Given that $\log 4586=3.6615$, find
(i) $\log 45.86$
(ii) $\log 45860$
(iii) $\log 0.4586$
(iv) $\log 0.004586$
(v) $\log 0.04586$
(vi) $\log 4.586$

Answer 1

(i) $\log (4586\times {10}^{-2})=-2+\log \left(4586\right)$

$=-2+3.6615=1.6615$

(ii) $\log (4586\times 10)=1+\log \left(4586\right)$

$=1+3.6615=4.6615$

(iii) $\log (4586\times {10}^{-4})=-4+\log \left(4586\right)$

$=-4+3.6615=0.3385$

(iv) $\log (4586\times {10}^{-6})=-6+\log \left(4586\right)$

$=-6+3.6615=-2.3385$

(v) $\log (4586\times {10}^{-5})=-5+\log \left(4586\right)=-1.3385$

(vi) $\log (4586\times {10}^{-3})=-3+\log \left(4580\right)=0.6615$