Solving a 'Harvard' University entrance exam question

Added: 2026-05-15

[00:00:00]Hello Friends find the value of 'n' If (-19)^n=19 let's have a solution so, we have a problem of (-19)^n=19 as -19=-1.19 (-1.19)^n=19 as we know (a.b)^n=a^n.b^n then (-1)^n.(19)^n=19 according to euler's identity e^(iπ)+1=0 then e^(iπ)=-1 so here, we can write it as (e^(iπ))^n.19^n=19 since (a^m)^n=a^mn we have e^(iπn).19^n=19 now, take natural logarithm (ln) on both sides ln(e^(iπn).19^n)=ln19 as ln(a.b)=lna+lnb apply this formula lne^(iπn)+ln19^n=ln19 as lnm^n=nlnm It will be iπnlne+nln19=ln19 since lne=1 now, we common 'n' n(iπ+ln19)=ln19 to find 'n', divide by (iπ+ln19) on both sides n(iπ+ln19)/(iπ+ln19)=ln19/(iπ+ln19) then iπ+ln19 can cancels n=ln19/(iπ+ln19) which is our solution