Is it possible to compute $-\log\left({\sqrt{1.8\times 10^{-5}\times 0.1}}\right)$ without a calculator? The following question is part of a chemistry problem that came in the Dhaka University admission exam 2013-14.


What is $-\log\left({\sqrt{1.8\times 10^{-5}\times 0.1}}\right)$?  
(a) 2.672  
(b) 2.772  
(c) 2.872  
(d) 2.972


My attempt
$$\begin{aligned} -\log_{10}\left({\sqrt{1.8\times 10^{-5}\times 0.1}}\right) &= -\log_{10}\left({\sqrt{1.8\times 10^{-6}}}\right)\\ &\approx -\log_{10}\left({\sqrt{\left(10^{-3}\right)^2}}\right) \\ &=-\log_{10} \left(10^{-3}\right) \\ &= 3 \end{aligned}$$
The options are cruelly close to one another, so the approximation that I have done is of no use to me. How do I find the correct answer by hand quickly (as this is a competitive exam)?
 A: Well, $(\sqrt{1.8})^8=$ $(1.8)^4=(3.24)^2$ which is just over $10$. So $8\log(\sqrt{1.8})$ has to be just slightly over $\log_{10} 10 =1$, which gives $\log_{10}(\sqrt{1.8})$ has to be just a bit over $1/8$.
Then $$-\log_{10}(\sqrt{1.8×10^{-6}})$$ $$=
\log_{10}\left(\frac{1}{\sqrt{1.8×10^{-6}}}\right)$$
$$=\log_{10}\left(\frac{10^3}{\sqrt{1.8}}\right)$$
$$\approx 3 - 1/8.$$
So (c).
A: We can quite easily determine, by hand, that $1.8^2 = 3.24$ and $3.24^2 = 10.4976$. This means $1.8^4 \approx 10$, so
$$4\log_{10}1.8 \approx 1 \; \; \to \; \; \log_{10}1.8 \approx 0.25 \tag{1}\label{eq1A}$$
Therefore,
$$\begin{equation}\begin{aligned}
-\log_{10}\left({\sqrt{1.8\times 10^{-6}}}\right) & = -\frac{1}{2}\left(\log_{10}1.8 - 6\right) \\
& \approx -\frac{1}{2}(-5.75) \\
& = 2.875
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Since $\log_{10}1.8$ is actually marginally larger than $0.25$, the final answer would be just a bit less than $2.875$. This makes it fairly clear the correct answer is (c), i.e., $2.872$.
A: Working with whole numbers
$$1.8\times 10^{-5}\times 0.1= \frac 9 5 \times 10^{-6}$$
$$\sqrt{1.8\times 10^{-5}\times 0.1}=\frac{3}{ \sqrt{5}}\times 10^{-3}$$
$$\frac{3}{ \sqrt{5}}=\frac{3}{ \sqrt{4+1}}=\frac 3 2 \times \frac 1{\sqrt{1+\frac 14}}\sim\frac 3 2 \times \left(1-\frac 18\right)=\frac{21}{16}=1+\frac 5 {16}$$
$$\log_{e}\left(1+\frac 5 {16}\right)\sim\frac 5 {16}\implies \log_{10}\left(1+\frac 5 {16}\right)\sim  \frac {5}{16 \times 2.30}  \sim 0.135$$
Then $-2.865$ by hand
Edit
A good rational approximation of $\log_e(10)$ is  $\frac{76}{33}$
$$\log_{10}\left(1+\frac 5 {16}\right)\sim  \frac {5}{16} \times  \frac{33}{76}=\frac{165}{1216}\sim\frac{165}{1215}=\frac{11}{81}\times  \frac{125}{125}=\frac{1375}{10125} \sim\frac{1375}{10000}=0.1375$$
A: The $\sqrt{1.8}$ isn't insignificant. in fact it is essential to the answer.
The sum approximates to $-\log_{10} 1.35 + 3$, and the answers are then effectively

*

*$10^{0.33}\approx 10^{\frac13}>2$

*$10^{0.23}\approx 10^{\frac14}> \sqrt{3}$

*$10^{0.13}\approx 10^{\frac18}\approx \sqrt{\sqrt{3}}\approx1.3$

*$10^{0.03}\approx 10^{\frac1{30}}<1.2$
which gives (c) as the answer.
A: Write (all logs are w.r.t base 10) $-\log(\sqrt{1.8}\cdot 10^{-3})=3-\log(1.8)/2$ using properties of the logarithm. Now, one sees  $1.8<2<10^{1/3}$. Also, by just multiplying one finds $1.8^4=(1.8^2)^2=10.24$, very close to $10$. Thus, $1/4 <\log(1.8) < 1/3$ such that $3-1/8 = 2.875 >3-\log(1.8)/2 >3-1/6 > 2.82$.
Thus, answer c) is correct.
