Change of Base & Reciprocal Log Property

We have :$$\frac{\log_{15}{a}+\log_{32}{a}}{(\log_{15}{a})(\log_{32}{a})}=4$$
We get $$\frac{\left(\frac{\log a}{\log\ 15}+\frac{\log a}{\log32}\right)}{\frac{\log a}{\log\ 15}\times\ \frac{\log a}{\log32}\ \ }=4$$
we get $$\log a\left(\log32\ +\log\ 15\right)=4\left(\log\ a\right)^2$$
we get $$\left(\log32\ +\log\ 15\right)=4\log a$$
=$$\log480=\log a^4$$
=$$a^4\ =480$$
so we can say a is between 4 and 5 .

It is given that  $$\log_{\sqrt{3}}{(x)}+\frac{\log_{x}{(25)}}{\log_{x}{(0.008)}}=\frac{16}{3}$$, which can be written as:

=> $$2\log_3x+\log_{0.008}25\ =\ \frac{16}{3}$$

=> $$2\log_3x+\log_{\frac{8}{1000}}25\ =\ \frac{16}{3}$$

=> $$2\log_3x+\log_{\frac{1}{125}}25\ =\ \frac{16}{3}$$

=> $$2\log_3x+\log_{5^{-3}}\left(5\right)^2\ =\ \frac{16}{3}$$

=> $$2\log_3x-\frac{2}{3}=\ \frac{16}{3}$$

=> $$2\log_3x=\frac{16}{3}+\frac{2}{3}$$

=> $$2\log_3x=6$$

=> $$\log_3x^2=6\ =>\ x^2\ =\ 3^6$$

Hence, $$\log_3\left(3\cdot x^2\right)\ =\ \log_3\left(3\cdot3^6\right)\ =\log_33^7\ =\ 7$$

The first term of the expression can be rewritten as $$\frac{\frac{1}{3}\log_2\left(a+b\right)}{\log_2c}$$
Using the property $$\frac{m}{n}\log_ab=\log_ab^{\frac{m}{n}}$$ this can be rewritten as

$$\frac{\log_2\left(a+b\right)^{\frac{1}{3}}}{\log_2c}$$

And finally using the property $$\frac{\log_ba}{\log_bc}=\log_ca$$, we can rewrite the expression as 

$$\log_c\left(a+b\right)^{\frac{1}{3}}$$

Doing identical operations in the second term, we get the entire left-hand side to be:

$$\log_c\left(a+b\right)^{\frac{1}{3}}+\log_c\left(a-b\right)^{\frac{1}{3}}$$
Using property $$\log_ca+\log_cb=\log_c\left(ab\right)$$ we get

$$\log_c\left[\left(a+b\right)^{\frac{1}{3}}\left(a-b\right)^{\frac{1}{3}}\right]$$
$$\log_c\left[\left(a+b\right)\left(a-b\right)\right]^{\frac{1}{3}}$$
$$\log_c\left[\left(a^2-b^2\right)\right]^{\frac{1}{3}}$$

This expression is given to be equal to 2/3
Using the definition of log: $$\log_cN=a\ $$ which is $$c^a=N$$
we get:$$c^{\frac{2}{3}}=\left(a^2-b^2\right)^{\frac{1}{3}}$$

Cubing both sides:
$$c^2=a^2-b^2$$
Finally giving $$a^2=b^2+c^2$$

We have upper limits on b and c as 10, and we want to maximize the value of a squared. 
This can be thought of as a right-angled triangle, and the value of a will be maximum when both b and c are equal to 10, giving $$a^2=200$$, but this would not give an integer value of a
We need to adjust $$a^2$$ to the biggest square less than 200, which is 196
Giving the value of $$a$$ as 14. 

Therefore, 14 is the correct answer.