2 votes 2 votes Please help with this question:- $(\sqrt{243}+3)^x+(\sqrt{243}-3)^x=15^x$. Quantitative Aptitude general-aptitude + – Devshree Dubey asked Aug 11, 2019 • edited Aug 11, 2019 by ankitgupta.1729 Devshree Dubey 983 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments ankitgupta.1729 commented Aug 12, 2019 reply Follow Share yes. https://www.wolframalpha.com/input/?i=(sqrt(243)%2B3)%5Ex+%2B+(sqrt(243)-3)%5Ex+-+15%5Ex 0 votes 0 votes Devshree Dubey commented Aug 14, 2019 reply Follow Share @ankitgupta.1729,Ideally if we go to see,in a way since here 'x' hasn't been restricted to Real Numbers or for that matter Natural Numbers you've assumed x for 'Real' values. Isn't it? 0 votes 0 votes ankitgupta.1729 commented Aug 14, 2019 reply Follow Share @Devshree Dubey Yeah, I have assumed that you were asking solution for real values of x..if there is no restriction then we have to take complex roots as mentioned in the above link which I guess difficult to find. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes After applying properties of log and doing all calculation I am x=-1.20 negative value and on above equation, if you put x=1 it will give 234/15 which is not equal to 1, sorry x=0 will not be a solution here I wrongly mentioned it STUDYGATE2019 answered Aug 14, 2019 • edited Aug 22, 2019 by STUDYGATE2019 STUDYGATE2019 comment Share Follow See 1 comment See all 1 1 comment reply Shaik Masthan commented Aug 17, 2019 reply Follow Share how x=0, can satisfy ? 2 votes 2 votes Please log in or register to add a comment.
0 votes 0 votes $(\frac{√243+3}{15})^x+(\frac{√243-3}{15})^x=1$ $√243+3 = 12.58/15 <1 , √243+3 = 18.58/15 >1$ $(\frac{√243+3}{15})^x$ will be >1 if X >1 $(\frac{√243-3}{15})^x$ will be <1 always so sum of these two cannot be 1 for any real positive number. smsubham answered Mar 22, 2020 smsubham comment Share Follow See all 0 reply Please log in or register to add a comment.