How do you solve #log x + log(x+21) =2#?

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Answer 1 · 2018-06-20T12:26:55

#x=4#

Here,

#logx+log(x+21)=2#

We assume that the common base of the log as 10.

#log_10x+log_10(x+21)=2#

#=>log_10(x*(x+21))=2to[becauselogM+logN=log(MN)#

#=>log_10(x^2+21x)=2#

#=>x^2+21x=10^2to[becauseX=log_aY<=>Y=a^X]#

#=>x^2+21x-100=0#

#=>x^2+25x-4x-100=0#

#=>x(x+25)-4(x+25)=0#

#=>(x-4)(x+25)=0#

#=>x=4 or x=-25#

But for #x=-25# log x is not defined.

Hence #x=4#

Check :

#LHS=log_10 4+log_10 (4+21)=log_10 4+log_10 25#

#:.LHS=log_10(4xx25)=log_10 100=log_10 10^2=2=RHS#