Quotient of Arguments to a Logarithm The problem I'm working on is: There are p , q >

gorgeousgen9487 2022-07-02 Answered
Quotient of Arguments to a Logarithm
The problem I'm working on is:
There are p , q > 0 satisfying log 9 p = log 12 q = log 16 ( p + q ). Find q / p
I set all of the logarithms equal to a variable t, so I could say
9 t = p , 12 t = q , 16 t = p + q
which implies
9 t + 12 t = 16 t .
I also know that
q p = 12 t 9 t = ( 4 3 ) t
p + q q = 16 t 12 t = ( 4 3 ) t
and therefore
q p = p + q q .
I haven't gotten any farther than this, so does anyone have advice?
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Answers (2)

Jayvion Tyler
Answered 2022-07-03 Author has 23 answers
We have 9 16 = 144 = 12 2 . Thus q 2 = p ( p + q ). Rewrite as
q 2 p q p 2 = 0.
Divide through by p 2 . We get
( q p ) 2 q p 1 = 0.
Solve this quadratic equation for q p , using the Quadratic Formula.
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Gretchen Schwartz
Answered 2022-07-04 Author has 5 answers
Notice that p + q q = p q + 1 = 1 q / p + 1. Substituting x = q p into your equation, you thus get
x = 1 x + 1
and multiplying through by x gives you x 2 = 1 + x . This is a quadratic equation, which you can solve for x (which is equal to q p .)
Keep in mind that it is specified that p , q > 0
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