Question

Find all the real solutions to the logarithmic equation
$\ln (x+1)+\ln \left(x\right)=\ln \left(2\right)$

• A. $1,-2$
• B. $0,\ln (-2)$
• C. $e,e^2$
• D. $-2$
• E. $1$

Answer 1

Answer: (E)

$1$

x> 0 and x>1

Let both sides be exponents of the base e. The equation $\ln (x+1)+\ln \left(x\right)=\ln \left(2\right)$can be rewritten as $e^{\ln (x+1)+\ln \left(x\right)}=e^{\ln \left(2\right)}$ or $e^{\ln \left(\right(x+1\left)\right(x\left)\right)}=e^{\ln \left(2\right)}$.

By now you should know that when the base of the exponent and the base of the logarithm are the same, the left side can be written $x$.

The equation $e^{\ln \left(\right(x+1\left)\right(x\left)\right)}=e^{\ln \left(2\right)}$ can now be written as $(x+1)\left(x\right)=2$ that is $x^2+x-2=0$.

Now finding the roots of the above quadratic equation as follows:

$x^2+x-2=0$

$x^2+2x-x-2=0$

$x(x+2)-1(x+2)=0$

$x+2=0$ and $x-1=0$

$x=-2$ and $x=1$

Since we have to find the real solution, therefore, $x=1$.

Hence, option E is correct.