Question

What is the first step in solving In(x - 1) = In6 - Inx for x?

Answer 1

Explanation:

The first step is to create the domain of this equation.

`\ln(x-1)=\ln6-\ln x\\\\D:\ x-1>0\ \wedge\ x>0\\\\x>1\ \wedge\ x>0\Rightarrow x>1`

`\ln(x-1)=\ln6-\ln x\qquad\text{use}\ \log_ab-\log_ac=\log_a(b)/(c)\\\\\ln(x-1)=\ln(6)/(x)\iff x-1=(6)/(x)\\\\(x-1)/(1)=(6)/(x)\qquad\text{cross multiply}\\\\x(x-1)=(1)(6)\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\\\(x)(x)+(x)(-1)=6\\\\x^2-x=6\qquad\text{subtract 6 from both sides}\\\\x^2-x-6=0\\\\x^2+2x-3x-6=0\\\\x(x+2)-3(x+2)=0\\\\(x+2)(x-3)=0\iff x+2=0\ \vee\ x-3=0\\\\x+2=0\qquad\text{subtract 2 from both sides}\\x=-2 \\otin D\\\\x-3=0\qquad\text{add 3 to both sides}\\x=3\in D`

Solution:

x = 3

Answer 2

The first step would to be use quotient rule.

3

Explanation:

ln(x-1)=ln(6)-ln(x)

The first step would to be use quotient rule there on the right hand side:

ln(x-1)=ln(6/x)

*Quotient rule says ln(a/b)=ln(a)-ln(b).

Now that since we have ln(c)=ln(d) then c must equal d, that is c=d.

ln(x-1)=ln(6/x)

implies

x-1=6/x

So you want to shove a 1 underneath the (x-1) and just cross multiply that might be easier.

`(x-1)/(1)=(6)/(x)`

Cross multiplying:

`x(x-1)=1(6)`

Multiplying/distribute[/tex]

`x^2-x=6`

Subtract 6 on both sides:

`x^2-x-6=0`

Now this is not too bad to factor since the coefficient of x^2 is 1. All you have to do is find two numbers that multiply to be -6 and add up to be -1.

These numbers are -3 and 2 since -3(2)=-6 and -3+2=-1.

So the factored form of our equation is

`(x-3)(x+2)=0`

This implies that x-3=0 or x+2=0.

So solving x-3=0 gives us x=3 (just added 3 on both sides).

So solve x+2=0 gives us x=-2 (just subtracted 2 on both sides).

We need to see if these are actually the solutions by plugging them in.

Just a heads up: You can't do log(negative number).

Checking x=3:

ln(3-1)=ln(6)-ln(3)

ln(2)=ln(6/3)

ln(2)=ln(2)

This is true.

Checking x=-2:

ln(-2-1)=ln(6)-ln(-2)

ln(-3)=ln(6)-ln(-2)

We don't need to go further -2 makes the inside of our logarithms negative above.

The only solution is 3.