Question

The value of x in the equation log(x-1)9=2 is

Answer 1

`D:x-1>0 \wedge x\\-1\\ot=1\\ D:x>1 \wedge x\\ot=2\\ D:x\in(1,2)\cup(2,\infty)\\\\ \log_(x-1)9=2\\ (x-1)^2=9\\ x-1=3 \vee x-1=-3\\ x=4 \vee x=-2\\-2\\ot \in D\\ \boxed{ x=4}`

Answer 2

`\log_((x-1)) 9=2`

the domain:

`x-1 >0 \ \land \ x-1 \\ot=1 \\ x>1 \ \land \ x \\ot= 2 \\ x \in (1; 2) \cup (2;+\infty)`

the equation:

`\log_((x-1))9=2 \\ (x-1)^2=9 \\ √((x-1)^2)=√(9) \\ |x-1|=3 \\ x-1=3 \ \lor \ x-1=-3 \\ x=4 \ \lor \ x=-2`

4 is in the domain
-2 is not in the domain

The answer:

`x=4`

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Also, the answer in 13 is {8}. -3 is not in the domain. Replace x with -3 and you'll see:

`x=√(5x+24) \\ -3=√(5 * (-3)+24) \\ -3=√(-15+24) \\ -3=√(9) \\ -3=3`
It's not true so -3 isn't a solution to this equation.