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 $x=(9^6+7)/3$ .

How do you solve 4-4 + log _ 9 (3x - 7) = 64-4 + log _ 9 (3x - 7) = 6?