Question

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

• A. $1$
• B. $e^2$
• C. $\frac{1}{(e^2-1)}$
• D. $\frac{1}{\left(\ln \right(2)-1)}$
• E. no real solutions

Answer 1

The correct option is C $\frac{1}{(e^2-1)}$

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

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 $e^{\ln \left(\right(x+1\left)/\right(x\left)\right)}=e^2$ can now be written as $\frac{(x+1)}{\left(x\right)}=e^2$ that is

$(x+1)=x{e}^2$

$\Rightarrow x{e}^2-x=1$

$\Rightarrow x(e^2-1)=1$

$\Rightarrow x=\frac{1}{(e^2-1)}$

Therefore, $x=\frac{1}{(e^2-1)}$.

Hence, option C is correct.