What is the solution to #2log_9(x) = log_9 8 +log_9(x-2)#?

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Answer 1 · 2018-01-23T10:37:16

#x=4#

we need the laws of logs
providing all the logs are to the same base

#log(XY)=logX+logY#

#log(X/Y)=logX-logY#

#logX^n=nlogX#

we are given

#2log_9x=log_9 8+log_9(x-2)#

using the above

#log_9x^2=log_9 8(x-2)#

because the bases are the same

#=>x^2=8(x-2)#

#=>x^2-8x+16=0#

factorising

#(x-4)^2=0#

#:.x=4#