CAT 2024

Slot 2

QA

Question & Solution

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 \(\log_c N = a\) which means \(c^a = N\). So we get: \(c^{\frac{2}{3}} = (a^2 - b^2)^{\frac{1}{3}}\)

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.