Question

(log x)/(ry-qz)=(log y)/(pz-rx)=(log z)/(qx-py)
, show that x^p*y^q*z^r
​

Answer 1

Step-by-step explanation:

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Vaskhar10

02.09.2019

Math

Secondary School

answered

If log x/ry-qz=log y/pz-rx=log z/qx-py

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ksjxjsj

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Step-by-step explanation:

I'm going to try it again the next one more than that and it will support it again KKK you to get

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THANKS

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PrachiVakshi

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Answer:

Step-by-step explanation:

log x/(ry-qz)=log y/(pz-rx)=log z/(qx-py)=k

log x =k(ry-qz)

log y=k(pz-rx)

log z=k(qx-py)

x^p*y^q*x^r=A

=> log x^p*y^q*z^r=log A

=>log x^p + log y^q + log z^r = log A

=>p log x + q log y + r log z = log A

=>pk(ry-qz) + qk(pz-rx) + rk(qx-py)=log A

k (pry-pqz+pqz-qrx+qrx-pry)=log A

=>k * 0=log A

=>A=10^0

=>A=0

.°. x^p * y^q * z^r=1

Hence Proved.

Hope you have satisfied with this answer.So please follow me and thank me and make me as brainlesset soon and vote my answer.

Answer 2

Question :

$\sf If \: \dfrac{ log(x) }{(ry - qz)} \: = \: \dfrac{ log(y) }{(pz - rx)} = \: \dfrac{ log(z) }{(qx - \: py)} \: , \: show \: that \: (x^p \times y^q \times z^r \: = \: 1)$

Given :

$\sf \bullet \: \: \dfrac{ log(x) }{(ry - qz)} \: = \: \dfrac{ log(y) }{(pz - rx)} = \: \dfrac{ log(z) }{(qx - \: py)}$

To prove :

$\sf \: x^p \times y^q \times z^r \: = \: 1$

Identity used :

$\sf \bullet \: log(ab) \: = \: log(a) + log(b) \\ \\ \sf \bullet \: \: log( \dfrac{a}{b} ) = log(a) - log(b) \\ \\ \sf \bullet \: \: log( {a}^{m} ) = m \times log(a) \\ \\ \sf \bullet \: log(1) = 0 \\ \\ \sf \bullet \: If \: \: log(a) = log(b) \: , \: then \: a = b$

Solution :

$\sf \: \: \dfrac{ log(x) }{(ry - qz)} \: = \: \dfrac{ log(y) }{(pz - rx)} = \: \dfrac{ log(z) }{(qx - \: py)} \: = \: k \: \{ \: let \} \\ \\ \sf \bullet \: \: \dfrac{ log(x) }{(ry - qz)} \: = k \\ \sf \implies \: log(x) = k(ry - qz) \\ \\\sf \: \bullet \: \: \dfrac{ log(y) }{(pz - rx)} = k\: \\ \sf \implies \: log(y) = \: k (pz - rx) \\ \\ \sf \bullet \: \dfrac{ log(z) }{(qx - \: py)} \: = \: k \\ \sf \implies \: log(z) = k(qx - py)$

Now. we need to prove

$\sf \: x^p \times y^q \times z^r \: = \: 1$

LHS ,

$\sf \: x^p \times y^q \times z^r \: = \: m \: \{ \: let \} \\ \\ \sf \: take \: log \: on \: both \: side \\ \\ \sf \implies \: log(\: x^p \times y^q \times z^r \: ) = log(m) \\ \\ \sf \implies \: log( {x}^{p} ) + log( {y}^{q} ) + log( {z}^{r} ) = log(m) \\ \\ \sf \implies \: \{p \times log(x) \} + \{ \: q \times log(y) \} \: + \: \{ \: r \times log(z) \} \: = log(m)$

Put value of log(x) , log(y) & log(z) in above equation.

$\sf \implies \: \{ \: p \times k(ry - qz) \} \: + \: \{ \: q \times \: k (pz \: - rx )\: \} \: + \{ \: r \times \: k \: (qx \: - \: py )\: \} = log(m ) \\ \\ \sf \implies \: \: k(pry \: - pqz) \: + \: k(qpz -qrx ) + k(rqx -rpy ) = log(m) \\ \\ \sf \implies \: k \{pry \: - pqz + qpz -qrx + \: rqx -rpy \} = log(m) \\ \\ \sf \implies \: k (0) = log(m) \\ \\ \sf \implies \: 0 \: = \: log(m) \\ \\ \sf \implies \: log(1) = log(m) \\ \\ \sf \implies \: m \: = \: 1$

Now , RHS = 1

This gives, RHS = LHS

Hence PROVED.