# Math Help - log change formula

1. ## log change formula

Can someone please help me with these optimisation questions finding it very difficult (ps have given what i think is valid answer to q1 but cant do q2 at all.)
note: 2^6 measn 2 to the power of 6,
note: theta is a capital theta
note: log c (a) means log to the base c of a

1) Let a, b, c be positive real numbers. Show that loga(b) = logc(b)/ logc(a).

( i think i have the answer to this can someone pls confirm this is a thoroguh enough proof

let y= log a (b)
then a^y =b
so taking logarithms to the base c gives

log c (a^y)= log c (b)
implying
ylog c(a) = log c (b)
so: y = log c (b)/log c (a)

2) Let a, b be real numbers, with a>0 and b > 0. Show (n + a)^b = theta(n^b). Show that (log n)^b = o(n^a). (For concreteness let’s take ‘log’ to mean ‘log2’.

thanx edgar

2. Originally Posted by edgar davids
Can someone please help me with these optimisation questions finding it very difficult (ps have given what i think is valid answer to q1 but cant do q2 at all.)
note: 2^6 measn 2 to the power of 6,
note: theta is a capital theta
note: log c (a) means log to the base c of a

1) Let a, b, c be positive real numbers. Show that loga(b) = logc(b)/ logc(a).
(It does not work if some are equal to 1).

$\log_a b = x, \ \log_c b = y , \ \log_c a = z$.
$a^x = b , \ c^y = b , \ c^z = a$
$a^x = b \implies (c^z)^x = b \implies c^{xz} = c^y \implies xz = y$.