What is x if $log_3 (x-1) + log_3 (x-9) = 2$ ?

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Answer 1 · 2018-04-04T18:36:27

$x=10$

We know that ,

$color(blue)((1)log_aM+log_aN=log_a(MN)$

$color(blue)((2)log_aX=n<=>X=a^n$

Here,

$log_3(x-1)+log_3(x-9)=2...toApply (1)$

$=>log_3((x-1)(x-9))=2...toApply (2)$

$=>(x-1)(x-9)=3^2$

$=>x^2-x-9x+9=9$

$=>x^2-10x=0$

$=>x(x-10)=0$

$=>x=0 or x-10=0$

$=>x=0 or x=10$

But, $x=0=>x-1=0-1=-1 < 0$

$and x-9=0-9=-9 < 0$

So, $x!=0$

i.e. $x=10$

What is x if log_3 (x-1) + log_3 (x-9) = 2log_3 (x-1) + log_3 (x-9) = 2?