How do you solve for x in #2log_3x=log_3 32+log_3 2#?

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Answer 1 · 2017-02-09T19:24:49

#8sqrt2#

For logarithms we have the following definition

#log_ab=c=>a^c=b---(1)#

we also have the standard results

#log_aX+log_aY=log_aXY##----(2)#

#log_aX-log_aY=log_a(X/Y)---(3)#

#log_aX^n=nlog_aX #color(white)(****)#---(4)#

so we have

#2log_3x=log_(3)32+log_(3)8#

using #" "(2)" on "RHS#

#2log_3x=log_(3)(32xx4)#

#2log_3x=log_(3)(128)#

#=>log_3x^2=log_(3)128#

#"using "(4)" on "LHS#

#because " the bases are the same we have:"#

#x^2=128#

#=>x=sqrt128#

#=>x=sqrt(64xx2)=8sqrt2#