How do you solve #3\cdot 14^ {-6g} = 26#?

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Answer 1 · 2017-04-02T10:36:30

#g = (ln 3 - ln 26)/(6 ln 14)#

Given:

#3*14^(-6g) = 26#

Divide both sides by #3# to get:

#14^(-6g) = 26/3#

Take logarithm base #14# both sides to get:

#-6g = log_14(26/3)#

Divide both sides by #-6# to get:

#g = -1/6 log_14(26/3) = 1/6 log_14(3/26)#

If you prefer logairthm in another base (e.g. #10# or #e#), then we can use the change of base formula:

#log_a b = (log_c b)/(log_c a)#

So:

#g = 1/6log_14(3/26) = (ln (3/26))/(6 ln 14) = (ln 3 - ln 26)/(6 ln 14)#