How do you solve #5^(5y-2)=2^(2y+1)#?

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Answer 1 · 2016-10-16T13:30:26

#5^(5y - 2) = 2^(2y + 1)#

#ln(5^(5y - 2)) = ln(2^(2y + 1))#

#(5y - 2)ln5 = (2y + 1)ln2#

#5yln5 - 2ln5 = 2yln2 + ln2#

#5yln5 - 2yln2 = ln2 + 2ln5#

#y(5ln5 - 2ln2) = ln2 + 2ln5#

Apply the rules #alnn = lnn^a#, #lna - lab = ln(a/b)# and #lna + lnb = ln(a xx b)# to simplify.

#y(ln781.25) = ln50#

#y = ln50/ln781.25#

#y = 0.59#

Hopefully this helps!