Why does #lna - lnb = ln(a/b)#?

Question

Isaiah Anthony · Accepted Answer

It does not matter what base we use providing the same base is used for all logarithms, here we are using bease #e#.

Let us define #A,B.C# as follows=:

# A = ln a iff a = e^A #,

# B = ln b iff b = e^B #

# C = ln (a/b) iff a/b = e^C #

From the last definition we have:

# a/b = e^C => e^C = (e^A)/(e^B) #

And using the law of indices:

# e^C = (e^A) (e^-B) = e^(A-B) #

And as as the exponential is a #1:1# monotonic continuous function, we have:

# C = A-B #

And so:

# ln (a/b) = ln a - ln b \ \ \ # QED

Abigail Allen · Answer

undefined The natural logarithm function ln has a property called the quotient rule, which states that ln(a) - ln(b) equals ln(a/b). This is because ln(a) - ln(b) simplifies to ln(a/b) due to the properties of logarithms.

Eva Adamson · Answer

undefined The natural logarithm of the quotient of two numbers, a and b, is equal to the difference of the natural logarithms of a and b. This property stems from the definition of logarithms and the properties of exponents. Specifically, the natural logarithm of a number is the exponent to which e (the base of the natural logarithm) must be raised to yield that number. Therefore, ln(a) represents the exponent to which e must be raised to equal a, and ln(b) represents the exponent to which e must be raised to equal b. When we subtract ln(b) from ln(a), we are essentially dividing the two exponents, resulting in ln(a/b), as the division of exponents corresponds to the quotient of their bases raised to the power of the difference of the exponents.