Prove the quotient law of logarithms by applying algebraic reasoning.?

Question

Prove the quotient law of logarithms by applying algebraic reasoning.?

Answer 1

1. The quotient law of logs is Log(A/B) = Log A - Log B
2. Let Log₁₀ A = x.  Then 10^x = A   {I'm using base 10, but any base could be used.  I will not use the subscript any further}
3. Similarly, let LogB = y.  Then 10^y = B
4. Substituting:  Log(A/B) = Log(10^x/10^y)
5. We know from algebra that 10^x/10^y = 10^x-y
6. There is also the log property that says log_axⁿ = n(log_a^x)
7. So Log(A/B) = Log(10x-y) = (x-y)(Log10)
1. Log10 = 1, so we are left with (x-y) in the right side of equation
2. And substituting for x-y yields LogA - LogB
3. Log(A/B) - LogA - LogB