2^p=3^q=48^r show that pq=r(4q+p)

Question

2^p=3^q=48^r

show that pq=r(4q+p)

Answer 1

2^p=3^q=48^r

Take ln on each side,

pln2 = qln3 = rln48

Solve p and q,

p = rln48/ln2

q = rln48/ln3

Therefore, pq = (rln48)^2/(ln2 ln3) = r[r(ln48)^2/(ln2 ln3)] ......(1)

But 4q+p = 4rln48/ln3 + rln48/ln2 = rln48(4/ln3 + 1/ln2) = rln48 (4ln2+ln3)/(ln2 ln3) = r(ln48)^2/(ln2 ln3) .....(2)

Plug (2) in (1),

pq = r(4q+p)

QED

 

Answer 2

Take a natural log, then ln(2^p) = ln(3^q) = ln(48^r),

                                   48 = 2^4 * 3, ln(48^r) = ln(2^4 *3 ) = 4ln2 + ln3

                Simplify,  p(ln2) =  q(ln3) = r( 4ln2 + ln3)

                For any real number to satisfy the first part of equations p(ln2) = q(ln3), p = ln 3,  q = ln 2,

Therefore,  p(ln 2) = q(ln3) = pq = r(4q + p)

                       A specific way to show that the equation pq = r(4q+p) by replacing a set of solution. 

                 

Answer 3

[Edit of original] Sorry, I have seen several questions with typos in one night, so wondered whether the first = was supposed to be a multiplication in a number theory question I thought you perhaps intended.

 

To eliminate ambiguity, it would have helped to add 5 words, "where these are real numbers" (you're quite right that one can't assume they are integers. Even more true that one can't assume the first equal sign is a "times", ha ha..I was just asking, and wondering since unfortunately many post are with typos, or missing parentheses, etc, so I jumped in with one guess as to what you meant. Funny how it would work with integers, "2pq" instead, and multiplication.)

 

Sorry again and glad others jumped in, yes, with real numbers you just take logs and use log(a^b) = b*log(a) in any log base including ln