Math Olympiad| Two methods were used| can you?

I used two different methods to achieve the solution to the math Olympiad question. #matholympiadproblem #maths

[@key_board_x]

a = Log[2](x)
a = Ln(x) / Ln(2)
Ln(2) = Ln(x) / a
Ln(x) = a.Ln(2)
b = Log[x](7)
b = Ln(7) / Ln(x)
Ln(7) = b.Ln(x)
Let's calculate: Ln(7) / Ln(2)
= Ln(7) / Ln(2) → recall: Ln(7) = b.Ln(x)
= b.Ln(x) / Ln(2) → recall: Ln(2) = Ln(x) / a
= b.Ln(x) / [Ln(x) / a]
= b.Ln(x) * [a / Ln(x)]
= ab.[Ln(x) / Ln(x)]
= ab
Log[14](x) = Ln(x) / Ln(14)
1 / Log[14](x) = Ln(14) / Ln(x)
1 / Log[14](x) = Ln(2 * 7) / Ln(x)
1 / Log[14](x) = [Ln(2) + Ln(7)] / Ln(x)
1 / Log[14](x) = [Ln(2) / Ln(x)] + [Ln(7) / Ln(x)] → recall: Ln(2) = Ln(x) / a
1 / Log[14](x) = [{Ln(x) / a} / Ln(x)] + [Ln(7) / Ln(x)]
1 / Log[14](x) = [Ln(x) / {a.Ln(x)}] + [Ln(7) / Ln(x)]
1 / Log[14](x) = [1/a] + [Ln(7) / Ln(x)] → recall: Ln(x) = a.Ln(2)
1 / Log[14](x) = [1/a] + [Ln(7) / a.Ln(2)]
1 / Log[14](x) = [1/a] + [(1/a) * Ln(7) / Ln(2)] → recall: Ln(7) / Ln(2) = ab
1 / Log[14](x) = [1/a] + [(1/a) * ab]
1 / Log[14](x) = [1/a] + [ab/a]
1 / Log[14](x) = [1 + ab]/a
Log[14](x) = a/(1 + ab)

[@RyanLewis-Johnson-wq6xs]

If a=Log[2,x], b=Log[x,7], Log[14,x]=a/(ab+1) final answer

[@JJONLINEMATHSCLASSchannel]

👍👍