what are the potential solutions of $ln(x^{2}-25)=0$?\n$x = pmsqrt{13}$\n$x=pm5$\n$x=pmsqrt{26}$\n$x=pm10$
what are the potential solutions of $ln(x^{2}-25)=0$?\n$x = pmsqrt{13}$\n$x=pm5$\n$x=pmsqrt{26}$\n$x=pm10$
Answer
Explanation:
Step1: Use the property of natural logarithm
Recall that if $\ln(a)=0$, then $a = e^0$. Since $e^0=1$, we have $x^{2}-25 = 1$.
Step2: Solve the quadratic equation
Add 25 to both sides of the equation $x^{2}-25 = 1$. We get $x^{2}=1 + 25$, so $x^{2}=26$. Then take the square - root of both sides, $x=\pm\sqrt{26}$.
Answer:
$x=\pm\sqrt{26}$