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.

Cross multiplying:

Multiplying/distribute[/tex]

Subtract 6 on both sides:

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

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.

## Answers ( )

Answer:The first step would to be use quotient rule.3

Step-by-step explanation:ln(x-1)=ln(6)-ln(x)

The first step would to be use quotient rulethere 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.

Cross multiplying:

Multiplying/distribute[/tex]

Subtract 6 on both sides:

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

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.

Step-by-step explanation:## The first step is to create the domain of this equation.

Solution:

## x = 3