Natural Logarithmic Equation ln (x+30) + ln x = ln 31

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rakesh05 | High School Teacher | (Level 1) Assistant Educator

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Given logarithmic equation is `ln(x+30)+lnx=ln31`  

        We know that `lnx+lny=ln(xy)`           

So,                        `ln((x+30)x)=ln31`     

As `lnx=lnyrArrx=y`  .

So,                    `(x+30)x=31`

or,                       `x^2+30x-31=0`

or,                      `x^2+31x-x-31=0`

or,                   `(x+31)(x-1)=0`

or,                     `x=1`  and `x=-31.`

As logarithm for negative numbers is not defined. We can discard x=-31.

So x=1  is the desired solution.

                

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The equation ln (x+30) + ln x = ln 31 has to be solved for x.

ln (x+30) + ln x = ln 31

Use the rule ln x + ln y  = ln(x*y)

ln((x+30)*x) = ln 31

=> (x+30)*x = 31

=> x^2 + 30x - 31 = 0

=> x^2 + 31x - x - 31 = 0

=> x(x + 31) - 1(x + 31) = 0

=> (x - 1)(x + 31) = 0

=> x = 1 and x = -31

As the logarithm of a negative number is not defined, eliminate x = -31

The solution of the equation is x = 1

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