Logarithims?

Question

(loga/p)=(logb/q)=(logc/r)=log...

All logarithims to the same base and x does not equal 1.

If (b^2)/(ac)=(x^y)

How do I find y in terms of p, q, and r?

Answer 1

log(f*g) = log(f) + log(g), log(f/g) = log(f) - log(g)
log(f*f) = log(f) + log(f) = 2log(f)

when you extend this you get

log(f^g) = glog(f)

So starting with

(b^2)/(ac)=(x^y)

Take logarithms

log[(b^2)/(ac)]=log[(x^y)]

log(b^2) - log(ac)=ylog(x)

2log(b) - log(a) - log(c) = ylog(x)

y = [2log(b) - log(a) - log(c)]/log(x)

Answer 2

since (b^2)/(ac)=(x^y)
So (b/a) * (b/c) = x^y
Taking log to base x
log(b/a) + log(b/c) =y
So
y = log b - log a + log b - log c = 2 log b - log a - log c (1)
Since loga/p)=(logb/q)=(logc/r)
log a - log b = log b - log q = log c - log r
log c = log b - log q + log r
log a = 2 log b - log q
From (1)
y= 2 log b - 2l og b + log q - log b + log q - log r
y =2 log q - log a - log r = log(q^2/ar)

Answer 3

(b^2)/(ac)=(x^y)
log(b^2/(ac)) = y log x
y = [log(b^2/(ac))]/log x
y = [logb^2 -log(ac)]/log x
y = [2logb - loga -log c]/log x
loga = log bp/q
log b = log cq/r
log c = log br/q
y = [2log(cq/r) - log(bp/q) -log (br/q)]/logx
y = log (c^2q^4/(r^3b^2))/logx