How do you solve $3 log_2 (x-1)=5 - log_2 (4)$ ?

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Answer 1 · 2015-12-28T16:05:23

It is $x=3$

It is

$3 log_2 (x-1)=5 - log_2 (4)=>log_2(x-1)^3+log_2 4=5=> log_2(4*(x-1)^3)=5=>4*(x-1)^3=2^5=>(x-1)^3=2^3=>x-1=2=>x=3$

Of course from the definition of the logarithm is

$x-1>0=x>1$

How do you solve 3 log_2 (x-1)=5 - log_2 (4) 3 log_2 (x-1)=5 - log_2 (4) ?