How do you solve #5=6^(3t-1)# ?

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Answer 1 · 2016-08-02T18:38:55

#t = 1/3((log 5)/(log 6)+1)#

Taking logs of both sides, we have:

#log 5 = log (6^(3t-1)) = (3t-1) log 6#

Divide both ends by #log 6# and transpose to get:

#3t - 1 = (log 5)/(log 6)#

Add #1# to both sides to get:

#3t = (log 5)/(log 6) + 1#

Divide both sides by #3# to get:

#t = 1/3((log 5)/(log 6)+1)#