Problem: logp+logq=log(p+q)\log p+\log q=\log (p+q)logp+logq=log(p+q) only if:
Answer Choices:
A. p=q=p=q=p=q= zero
B. p=q21−qp=\dfrac{q^{2}}{1-q}p=1−qq2
C. p=q=1p=q=1p=q=1
D. p=qq−1p=\dfrac{q}{q-1}p=q−1q
E. p=qq+1p=\dfrac{q}{q+1}p=q+1q Solution:
logp+logq=log(pq)\log p+\log q=\log (p q)logp+logq=log(pq). Thus we must have pq=p+qp q=p+qpq=p+q.
∴p=q/(q−1)\therefore p=q /(q-1) ∴p=q/(q−1)