1aditya can you please tell me how you got this above equation?......
\hspace{-16}$If $\mathbf{p,q,r>0\;}$,Then find $\mathbf{p,q,r}$ in \\\\ $\mathbf{\ln\left(pqr\right)=-2}$\\\\ $\mathbf{ln(p).\ln(q).\ln(r)=2}$\\\\ $\mathbf{\ln(p).\ln(q)+\ln(q).\ln(r)+\ln(r).\ln(p)=-1}$
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5 Answers
12.303 log pqr i= -2.
log pqr= -2/2.303
pqr is antilog -2/2.303.
from the second equation, ln(p+q+r)= 2.
so p+q+r is antilog 2/2.303
can we not solve the equations to fin the values?????????? the third one will have to be thought over...
21Just replace log p=a,log q=b...
Solve the 3 equations.
Ans:(p,q,r)=(e,1/e,1/e^2),(1/e,e,1/e^2)
262hint for others,
consider the eqn x3 - (-2)x2 +(-1)x -(2) =0 whose roots are ln(p),ln(q) and ln(r)
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1708\hspace{-16}$Here I am Discribing Aditiya Solution::$\\\\ $Let $\ln(a)=\alpha\;,\ln(b)=\beta\;,\ln(c)=\gamma$\\\\ So equation Convert into\\\\ $\ln(p.q.r)=-2\Leftrightarrow \ln(p)+\ln(q)+\ln(r)=-2$\\\\ So $\alpha+\beta+\gamma=-2$\\\\ $\alpha.\beta+\beta.\gamma+\gamma.\alpha=-1$\\\\ $\alpha.\beta.\gamma=2$\\\\ So we have to form a cubic equation whose Roots are $\alpha,\beta,\gamma$ is\\\\ $x^3-(\alpha+\beta+\gamma).x^2+(\alpha.\beta+\beta.\gamma+\gamma.\alpha).x-\alpha.\beta.\gamma=0$\\\\ $x^3+2x^2-x-2=0$