We define the logarithm base b of a (written log_b a) lo be the power X such tha

Question

We define the logarithm base b of a (written log_b a) lo be the power X such that X = log_b a if and only if b^x = a You can use this definition and real number exponent laws to prove many things. For instance, let's prove that log_b a^2 = 2log_b a. By the definition (by considering X = right hand side), this equation is equivalent to checking that b^2 log_ = a^2. We know b^2 log_b = (b^log_)^2 (*) by exponent laws. And we farther known directly by the definition of logs, that b^log a = a Thus the right hand side of (*) equals a^2 by substitution, so we have proved log_b a^2 = 2 log_b a. QED. In a similar way, using this definition and exponent laws only, prove the following equations assuming that every term is well-defined. log_c (ab) = log_c a + log_c b log_c (a^) = p log_C a (log_c (a)) (log_a(b)) = log_c b then conclude log_a = (b) = log_c(b)/log_c(a) No points given unless your solution clearly checks one side of each equation satisfying a defining condition of a relevant logarithm. Since this will prove the log laws, you better not use log laws!

Explanation / Answer

(i) logc (ab) = logc a + logc b

This is equivalent to checking

clogc a + logc b = ab

=> clogc a + logc b = clogc a clogc b (By exponent laws)

Since clogca = a and clogcb = b

=> clogc a + logc b = a * b = ab

(ii) logc (ap) = p logc a

This is equivalent to checking ap = cp logca

cp logca = (clogca)p   (Using exponent laws)

Since clogca = a

=> cp logca = ap

(iii) (logc a) (loga b) = (logc b)

This is equivalent to proving,

b = c(logc a) (loga b)

c(logc a) (loga b) = (c(logc a))(loga b) (by exponent laws)

But c(logc a) = a

=> c(logc a) (loga b) = a(loga b)

But a(loga b) = b

=> c(logc a) (loga b) = b

So (logc a) (loga b) = (logc b)

or loga b = (logc b) / (logc a)

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