Here is a much shorter solution: The function f(x)=(4+sqrt(15))^x + (4-sqrt(15))^x is monotonic increasing on [0,oo). Since (4+sqrt(15))(4-sqrt(15))=1 the function is symmetric f(x)=f(-x). ==> there are at most two solution f(x)=62. x=2 and x=-2 are two obvious solutions.
The plan is much more important than the detailed solution. We observe the two terms are reciprocals. The rest is a slog.
3,840の割算の最後は0ではなく1です。
Here is a much shorter solution: The function f(x)=(4+sqrt(15))^x + (4-sqrt(15))^x is monotonic increasing on [0,oo). Since (4+sqrt(15))(4-sqrt(15))=1 the function is symmetric f(x)=f(-x). ==> there are at most two solution f(x)=62. x=2 and x=-2 are two obvious solutions.
Thank you, you explain very well
8:31 >> 5÷5 = 1 instead 0
அருமை சகோதரி
Good one!
Под корнем надо быдо разложить на множители разность квадратов
Font cover the dum while writing we cant see it
х=2