$\ln(x+3)^{\frac{1}{2}} + \ln (4x-3)^{\frac{1}{2}} = \ln (5)$
So I understand that in order to solve this log function, I would have to square the square roots to simplify the equation.
But how does the number $e$ come into play?
$\ln(x+3)^{\frac{1}{2}} + \ln (4x-3)^{\frac{1}{2}} = \ln (5)$
So I understand that in order to solve this log function, I would have to square the square roots to simplify the equation.
But how does the number $e$ come into play?
No you dont have to square the square roots.
$ln(A)+ln(B)=ln(C)$
But we know that $ln(A)+ln(B)=ln(AB)$
So:
$ln(AB)=ln(C)$
$e^{AB}=e^{C}$ (Note: This is where e comes into play)
But we know that:
$a^b=a^c$ $b=c$
Therefore we can say that :
$e^{AB}=e^{C}$
$(AB)=C$