1 1 vote If $x$ and $y$ are positive real numbers such that $\log _{x}\left(x^{2}+12\right)=4$ and $3 \log _{y} x=1$, then $x+y$ equals $11$ $20$ $10$ $68$ Quantitative Aptitude cat2023-set1 quantitative-aptitude logarithms numerical-answer + – admin 5.3k points 1.5k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
1 1 vote By hit and trail x = 2 and y = 8 3*LogY^2 = 1 Log y^8 = 1 8 = y so x+y = 2+8 = 10 akash_kumar 9 answered Oct 20, 2025 akash_kumar 9 116 points comment Share Follow 0 reply Please log in or register to add a comment.