Question

Find all the real solutions to the logarithmic equation
$\ln (x^2-1)-\ln (x-1)=\ln \left(4\right)$

• A. $\frac{[1+\sqrt{(}17)]}{2},\frac{[1-\sqrt{(}17)]}{2}$
• B. $-2,3$
• C. $3$
• D. $\ln \left(3\right)$
• E. no real solutions

Answer 1

The correct option is C $3$

Let both sides be exponents of the base e. The equation $\ln (x^2-1)-\ln (x-1)=\ln \left(4\right)$can be rewritten as $e^{\ln (x^2-1)-\ln (x-1)}=e^{\ln \left(4\right)}$ or $e^{\ln \left(\right(x^2-1\left)/\right(x-1\left)\right)}=e^{\ln \left(4\right)}$.

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 $e^{\ln \left(\right(x^2-1\left)/\right(x-1\left)\right)}=e^{\ln \left(4\right)}$ can now be written as $(x^2-1)/(x-1)=4$ that is $(x-1)(x+1)/(x-1)=4$ or $x+1=4$ implies $x=3$.

Therefore, $x=3$.

Hence option C is correct.