can the power property of logarithms be derived from the power property of exponents using the equation bx = m? if not, explain why. if so show the derivation

Answer 1

Hi, Michael:

The simple answer is Yes!

By definition of a logarithm, if b^x = m, then x = log_b(m)

Now, suppose p = (b^x) x (b^y)

=> p = b^x+y [By the law of exponents]

Therefore, from the definition: x+y = log_b(p) . . . . . . . . . . . . . . . . .. . . . . . . . . . eqn (1)

Let b^x = m and b^y = n . This means that x = log_b(m) and y = log_b(n)

Substitute into eqn (1): log_b(m) + log_b(n) = log_b(p) [so multiplication becomes addition of logs]

Suppose q = (b^x)^y

=> q = b^xy [By the law of exponents]

=> xy = log_b(q) [Definition of log]

Therefore: y log_b(x) = log_b(q) [The power law for logs: an exponents multiplies the log]