For $a0,\ a\ne1$ and $M,N0$: \(\log_a\!\left(\dfrac{M}{N}\right)=\log_a M-\log_a N\

Question

For $a>0,\ a\ne1$ and $M,N>0$: $\log_a\!\left(\dfrac{M}{N}\right)=\log_a M-\log_a N$ Examples ExpressionSimplified form $\ln2-\ln5$ $\log_3 2-\log_3 20$ $\ln\dfrac{x}{x+4},\ x>0$ Take note The bases must match. Do not split $\log_a(M-N)$.

Accepted Answer

$\log_a b-\log_a c=\log_a\left(\dfrac bc\right)$

Expression	Simplified form
\(\ln2-\ln5\)
\(\log_3 2-\log_3 20\)
\(\ln\dfrac{x}{x+4},\ x>0\)

Quotient Law and Its Proof

Quotient Law and Its Proof

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