Answer:

log 199.1=2.2991

i.e. (iii) option is correct

Step:

antilog(0.2991)=1.991

∴log(1.991)=0.2991

log199.1=log(1.991×100)

             =log(1.991)+log(10²)…(loga+logb=log(ab))

             =log(1.991)+2log(10)…(nlogab=logabn)

As we know, loga(a)=1

log1.991=0.2991+2=2.2991

Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given that

[tex]\rm \: antilog(0.2991) = 1.991 \\ [/tex]

[tex]\rm\implies \:log(1.991) = 0.2991 \\ [/tex]

Now, Consider

[tex]\rm \: log(199.1) \\ [/tex]

can be rewritten as

[tex]\rm \: = \: log(1.991 \times 100) \\ [/tex]

We know,

[tex]\boxed{ \rm{ \:log(xy) = logx + logy \: }} \\ [/tex]

So, using this result, we get

[tex]\rm \: = \: log(1.991) + log(100) \\ [/tex]

[tex]\rm \: = \: 0.2991 + log( {10}^{2} ) \\ [/tex]

[tex]\rm \: = \: 0.2991 + 2log(10) \\ [/tex]

[tex]\rm \: = \: 0.2991 + 2 \times 1 \\ [/tex]

[tex]\rm \: = \: 0.2991 + 2 \\ [/tex]

[tex]\rm \: = \: 2.2991 \\ [/tex]

Hence,

[tex] \red{\rm\implies \:\rm \:log(199.1) = \: 2.2991} \\ [/tex]

So, option (iii) is correct.

[tex]\rule{190pt}{2pt}[/tex]

Additional Information :-

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ log_{x}(x) = 1}\\ \\ \bigstar \: \bf{ log_{x}( {x}^{y} ) = y}\\ \\ \bigstar \: \bf{ log_{ {x}^{z} }( {x}^{w} ) = \dfrac{w}{z} }\\ \\ \bigstar \: \bf{ log_{a}(b) = \dfrac{logb}{loga} }\\ \\ \bigstar \: \bf{ {e}^{logx} = x}\\ \\ \bigstar \: \bf{ {e}^{ylogx} = {x}^{y}}\\ \\ \bigstar \: \bf{log1 = 0}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]