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

Question

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

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Gianna 2 weeks 2021-09-10T15:10:01+00:00 2 Answers 0

Answers ( )

    0
    2021-09-10T15:11:22+00:00

    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 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.

    \frac{x-1}{1}=\frac{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.

    0
    2021-09-10T15:11:45+00:00

    Step-by-step 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\dfrac{b}{c}\\\\\ln(x-1)=\ln\dfrac{6}{x}\iff x-1=\dfrac{6}{x}\\\\\dfrac{x-1}{1}=\dfrac{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 \notin D\\\\x-3=0\qquad\text{add 3 to both sides}\\x=3\in D

    Solution:

    x = 3

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