Solving Exponential Equations using Logarithms | Math Olympiad

[@mohammedsaysrashid3587]

Thanks for sharing it was a wonderful introduction and explanation Sir 🙏....finally x=64

[@kassuskassus6263]

At the end, after some manipulations, we'll get: logx (log5+log3)/3=log15^2, so logx(log5+log3)/3=2(log5+log3), we simplify (log5+log3), we have logx=3*2, so logx=6, and then x=2^6=64.

[@Rt2026-y8n]

64 solve in 2 min

[@honestadministrator]

Denoting
log (z) to the base 2 as lg (z), one gets
log ( x ^ (1/3)) to the base 2
= 1/3 * lg (x)
log (3) to the base ( 2 √2)
= (1/3) lg (3)
Again a = 2 ^ lg (a)
Hereby
5 ^ ( (1/3) (lg (x)) * (x^ ((1/3) lg (3))
= (15) ^2
2^(1/3) *(lg (5)*lg (x) + lg (x)*lg (3))
= 15 ^ 2
Hereby
2 ^ ((1/3) lg ( x) lg (15)) = 15 ^ 2
15 ^ ( lg (x) /3) = 15.^ 2
lg (x) = 6
x = 2 ^ 6 . = 64

[@Quest3669]

Eqt well, x=64

[@gnanadesikansenthilnathan6750]

Understood but why log base 2√3 ^3 ?

[@Fjfurufjdfjd]

Χ=64

[@RajeshKumar-wu7ox]

64