How do you solve #\log _ { 3} ( 7t + 3) - \log _ { 3} t = \log _ { 3} 8#?

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Answer 1 · 2017-08-15T11:28:21

#t=3#

If #log_b(mn) = log_b(m) + log_b(n)# and

#log_b(m/n) = log_b(m) – log_b(n)#

So we can write the equation as #log_3((7t+3)/t)=log_(3)8#

#log_3((7t+3)/t) - log_(3)8 = 0#

#log_3((7t+3)/(t*8))=0#

If #log_(a)b=0#, then #b=1#

So #(7t+3)/(8t)=1#

#7t+3=8t#

giving #t=3#