Answer:
[tex]\log \:_{10}\left(\frac{1}{2}\right)=-\log \:_{10}\left(2\right)=-0.30102[/tex]
Thus, option A is true.
Step-by-step explanation:
Given the expression
[tex]log\left(\frac{1}{2}\right)[/tex]
[tex]=\log _{10}\left(2^{-1}\right)[/tex]
[tex]\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right),\:\quad \:x>0[/tex]
[tex]\log _{10}\left(2^{-1}\right)=-1\cdot \log _{10}\left(2\right)[/tex]
so the expression becomes
[tex]=-1\cdot \log _{10}\left(2\right)[/tex]
[tex]\mathrm{Multiply:}\:1\cdot \log _{10}\left(2\right)=\log _{10}\left(2\right)[/tex]
[tex]=-\log _{10}\left(2\right)[/tex]
substituting the value of log 2 = 0.30103
[tex]\:=-0.30102[/tex]
Thus,
[tex]\log \:_{10}\left(\frac{1}{2}\right)=-\log \:_{10}\left(2\right)=-0.30103[/tex]
Therefore, option A is true.
