Find the value of x in # sqrt(1+log_x(27))*log_3(x)+1=0#?

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Answer 1 · 2018-07-17T16:17:10

#sqrt(1+log_x(27))*log_3(x)+1=0#
#=>sqrt(log_3x*log_3x+3log_x(3)log_3(x)log_3x)=-1#

#=>log_3x*log_3x+3log_x(3)log_3(x)log_3x=1#

#=>(log_3x)^2+3log_3x-1=0#

#=>y^2+3y-1=0#

#=>y=(-3pmsqrt(9+4))/2#

#=>log_3x=(-3pmsqrt(9+4))/2#

So #log_3x=0.3028#

#=>x=3^(0.3028)~~1.4#

Again
#log_3x=-3.3#

#=>x=3^(-3.3)~~0.0265#