How do we solve 6^3t-1=5 for #t#?

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Answer 1 · 2016-06-05T07:09:24

#t=1/3log_6(30)#

In the absence of proper formatting I believe it is

#6^(3t-1)=5#. If so

#log_(6)5=3t-1# or

#3t=log_(6)5+1=log_(6)5+log_(6)6=log_(6)(5xx6)#

#3t=log_6(30)#

Hence, #t=1/3log_6(30)#

Answer 2 · 2016-06-05T08:38:56

For #6^(3t)-1=5#, the answer is #t = 1/3#.
For #6^(3t-1)=5#, the answer is #t=(1+log 5/log 6)/3#.

If it is #6^(3t)-1=5# then #6^(3t)=6#.

So, #3t = 1 and t = 1/3#.

For #6^(3t-1)=5#, equate logarithms.

#(3t-1) log 6 = log 5#. Solving,

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