How do you solve #4-4 + log _ 9 (3x - 7) = 6#?

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Answer 1 · 2015-12-29T08:16:00

#x=(9^6+7)/3#

#color(red)(cancel(color(black)(4)))-color(red)(cancel(color(black)(4)))+log_9(3x-7)=6#

#=>log_9(3x-7)=6#

To undo the logarithm with base #9#, exponentiate each side.

#=>9^(log_9(3x-7))=9^6#

#=>3x-7=9^6=531441#

#=>3x=531448#

#=>x=531448/3=177149 1/3#

This can also be written as #x=(9^6+7)/3#.