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#