Expert: Scott A Wilson - 1/6/2010

Question
Q1.If a^p=b,b^q=c,c^r=a then find the value of pqr
Q2.If a^p=b^q=c^r and b^2=ac then find the value of q(p+r)/pr
Q3.If 2^a=3^b=6^-c then find the value of 1/a+1/b+1/c


Answer
Q1. Note that by taking ln() of te equations, we get
p•ln(a) = ln(b), q•ln(b) = ln(c), and r•ln(c) = ln(a).

Using these, we can solve for p, q, and r.
They are p = ln(b)/ln(a), q = ln(c)/ln(b), and r =  ln(a)/ln(c).

Take these and put them into pqr to see what it is.
We get (ln(b)/ln(a))(ln(c)/ln(b))(ln(a)/ln(c)).

As can be seen, ln(a), ln(b), and ln(c) all occur once in the numerator and once in the denominator.  Since this happens, they can all be cancelled and the result is 1.


Q2. a^p = b^q = c^r and b² = ac.  Find q(p+r)/(pr).
Let K = p*ln(a) = q*ln(b) = r*ln(c).

From here, we can say that p = K/ln(a), q = K/ln(b), and r = K/ln(c).
It should be noted that since b² = ac, the ln() can be taken, so ln(b²) = ln(a) + ln(c).
Note that ln(b²) = 2*ln(b).

Putting p, q, and r into  q(p+r)/(pr) gives us [K/ln(b)][K/ln(a) + K/ln(c)]/{[K/ln(a)][K/ln(c)]}.

Factoring out K gives us K²[1/ln(b)][(1/ln(a) + 1/ln(c)]/{K²[1/ln(a)][1/ln(c)]},
and the K² can be cancelled, leaving [1/ln(b)][(1/ln(a) + 1/ln(c)]/[1/ln(a)][1/ln(c)].

Now if we look at the middle part, [(1/ln(a) + 1/ln(c)], the fractions can be combined.
This gives [ln(c) + ln(a)]/[ln(a)ln(c)].  Putting this in cancels out the ln(a)ln(c),
since both the top and the bottom are divided by it.

This leave [1/ln(b)][ln(a) + ln(c)].

Now from a while ago, it was stated that ln(a) + ln(c) = 2*ln(b).
What this comes down to then is [1/ln(b)][2*ln(b)].

The ln(b)'s cancel out, leaving us with 2.


Q3. Since 2^a = 3^b = 1/6^c, let K = a*ln(2) = b*ln(3) = -c*ln(6).

In this way, we can say that 1/a = ln(2)/K, 1/b = ln(3)/K, and 1/c = -ln(6)/K.

It is also known that ln(6) = ln(2*3) = ln(2) + ln(3).

In this way, we can say that 1/a + 1/b + 1/c = ln(2)/K + ln(3)/K - ln(6)/K.

Combining the fractions gives (ln(2) + ln(3) - ln(6))/K.

Looking at the numerator, is was stated that ln(6) = ln(2) + ln(3),
so putting that into the numerator gives ln(2) + ln(3) - ln(2) - ln(3) = 0.
With that, it doesn't matter what is in the denominator, for the number is 0.