Question

if px-1=qr,qy-1=rp,rz-1=pq then the value of 1/x +1/y+1/z is equal to _______ please solve it correctly please answer or i will report you

Answer 1

Answer:

pls do add

Step-by-step explanation:

1 /x +1/a

1/5

it will be helpful

Answer 2

Answer:

the question is

$if \: \: {p}^{x - 1} = qr \: \\ \: {q}^{y - 1} = rp \: \: \\ {r}^{z - 1} = pq \\ then \: find \: the \: value \: of \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \: ? \\ \\ the \: value \: of \\ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \\ \: \frac{1}{log \: pqr} (log \: p \: + log \: q \: + log \: r)$

Step-by-step explanation:

$given \: that \: {p}^{x - 1} = qr \: \\ \: {q}^{y - 1} = rp \: \: \\ {r}^{z - 1} = pq \\ then \: \: {p}^{x - 1} = qr \: \\ \frac{ {p}^{x } }{p} = qr \\ {p}^{x} =p qr \:$

taking logarithm on both sides

$log({p}^{x }) = log({pqr}) \\ x \: log(p) = \: log(pqr) \\ x = \frac{\: log(pqr)}{log(p)} \\ \frac{1}{x} = \frac{{log(p)} }{{\: log(pqr)}}$

then ,similarly we have

$\frac{1}{y} = \frac{{log(q)} }{{\: log(pqr)}} \\ \frac{1}{z} = \frac{{log(r)} }{{\: log(pqr)}}$

then , value of

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{{log(p)} }{{\: log(pqr)}} + \frac{{log(q)} }{{\: log(pqr)}} + \frac{{log(r)} }{{\: log(pqr)}}$

that is on rearranging,

$= \frac{{1} }{{\: log(pqr)}} ( log(p) + log(q) + log(r) )$

DEFINITIONS

here,we have use the concepts of logarithm and powers for simplification ,

some of them are

• log(bª)=a log b
• log(ab)=log a +log b
• log(a/b)=log a -log b

and

${p}^{x - y} = \frac{ {p}^{x} }{ {p}^{y} }$

${p}^{x - y} = \frac{ {p}^{x} }{ {p}^{y} }$

${ ({p}^{x}) }^{y} = { {p}^{x} }^{y}$

#SPJ3