How do you solve $Log_7(9-x)=2$ ?

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Answer 1 · 2015-11-28T03:13:26

I found answer: $x= -40$

To solve logarithmic equation, we switch it to exponential form

$log_by= x$ $harr$ $b^x = y$

$log_7(9-x)= 2$

$(9-x)= 7^2$ ; change to exponential form

$9-x= 49$ ;

$-x = 40$ $=> x= -40$

Check:

$log_7(9-(-40) =2$

$log_7(49)= 2$

$2= 2$

How do you solve Log_7(9-x)=2Log_7(9-x)=2?