A Very Nice Olympiad Math Exponential Problem l find value of x?

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[hangthuy]

log25^X=log25×10
log25^X=log25+log10
X=log25/log25+log10/log25
X=1+1/log25
X=1+1/2log5

[andredcavalcante]

25^x = 5^2.x = 250 = 25 . 10 = 5^2 . 10
log in b/s:
log 5^2.x = 2.x.log 5 = log 5^2.10 = log 5^2 + log 10 = 2.log 5 + 1
then
x = (2.log 5 + 1)/2.log 5
Finally:
x = 1 + 1/(2.log 5)
Less than 5 minutes in the worst formula editor ever.