Question

How do you solve $\ln x+\ln \left( x+5 \right)=\ln 14$?

Answer 1

To solve the equation in the logarithm given above we are going to use the following property of logarithm which states that $\log a+\log b=\log \left( ab \right)$. This property is valid for any base of the log. Using this property makes the problem easy to solve. Then using this property, you will get a quadratic equation in x and the solution of this equation is the solution of this quadratic equation.

Complete step-by-step solution:
The equation given in the above problem which we are asked to solve is as follows:
$\ln x+\ln \left( x+5 \right)=\ln 14$
In the L.H.S of the above equation, you can see that logarithm is written in the following way:
$\log a+\log b$
And we know that there is a property when two logarithms are added into each other we get,
$\Rightarrow \log a+\log b=\log \left( ab \right)$
Using the above property in the L.H.S of the given equation we get,
$\Rightarrow \ln \left( x\left( x+5 \right) \right)=\ln \left( 14 \right)$
Now, taking exponential on both the sides then we get,
$\Rightarrow {{e}^{\ln \left( x\left( x+5 \right) \right)}}={{e}^{\ln 14}}$
We know that:
${{e}^{\ln x}}=x$
Using the above property in ${{e}^{\ln \left( x\left( x+5 \right) \right)}}={{e}^{\ln 14}}$ we get,
$\Rightarrow x\left( x+5 \right)=14$
Multiplying x with $x+5$ in the above equation we get,
$\begin{aligned} & \Rightarrow {{x}^{2}}+5x=14 \\\ & \Rightarrow {{x}^{2}}+5x-14=0 \\\ \end{aligned}$
To solve the above quadratic equation, we are going to use the factorization method by writing the factors of 14 and then add or subtract them in such a way so that we can get 5.
Factors of 14 are as follows:
$\begin{aligned} & 14=7\times 2 \\\ & 14=14\times 1 \\\ \end{aligned}$
Now, if we use the first factors of 14 and subtract those two factors (7 and 2) then we get 5 so replacing 5 by (7 - 2) in the above quadratic equation in x we get,
${{x}^{2}}+\left( 7-2 \right)x-14=0$
Multiplying x by 7 and then 2 we get,
$\Rightarrow {{x}^{2}}+7x-2x-14=0$
Now, taking x as common from the first two terms and -2 as common from the last two terms we get,
$\Rightarrow x\left( x+7 \right)-2\left( x+7 \right)=0$
Taking $\left( x+7 \right)$ as common from the above equation we get,
$\Rightarrow \left( x+7 \right)\left( x-2 \right)=0$
Equating each bracket in the above equation to 0 we get,
$\begin{aligned} & x+7=0 \\\ & \Rightarrow x=-7; \\\ & x-2=0 \\\ & \Rightarrow x=2 \\\ \end{aligned}$
From the above solution, we are getting two values of x i.e. 2 and -7 in which -7 cannot be a solution because when we put x as -7 in the original equation then we get,
$\Rightarrow \ln \left( -7 \right)+\ln \left( -7+5 \right)=\ln 14$
Now, as you can see that the value written inside the log is negative which cannot be possible because log never take negative values.
Hence, only one solution is possible i.e. $x$ equal to 2.

Note: The mistake that you can do in the above problem is by accepting both the solutions of x which we got from solving the quadratic equation so make sure you won’t forget to use the property of the logarithm that log never takes negative values while checking the solutions of x. Also, the take home message from this problem is that we should always check the solutions of x which we are getting by putting those solutions in the original equation and then see whether they are fulfilling the conditions to satisfy that equation.