How do you solve $log(x)+log(x+3)=1$ ?

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Answer 1 · 2018-03-06T19:05:25

$x=2$

taking $" "logx-=log_10x$

using the law of logs

$logA+logB=logAB$

we have

$logx+log(x+3)=1$

$logx(x+3)=1--(1)$

definition of logs

$log_ab=c=>a^c=b$

$(1)rarr10^1=x(x+3)$

$:.x^2+3x=10$

$x^2+3x-10=0$

factorising and solving

$(x+5)(x-2)=0$

$=>x=-5, " or "x=2$

$because logx" is undefined for "x<=0$

$x=-5 " is not a solution"$

$x=2$

How do you solve log(x)+log(x+3)=1log(x)+log(x+3)=1?