Math, asked by Prabal5499, 9 months ago

if a^x=bc b^y=ca and c^z=ab prove that x/(x+1)+y/(y+1)+z/(z+1)=2 (WITHOUT LOGARITHM)
PLZ ANSWER CLEANLY AND IN DESCRIBATION​

Answers

Answered by xBrainlyKingXx
86

\rule{200}3

\color{red}\huge{\underline{\underline{\mathfrak{Question:-}}}}

if a^x=bc b^y=ca and c^z=ab prove that x/(x+1)+y/(y+1)+z/(z+1)=2

Solve without using logarithm

\rule{200}3

\color{blue}\huge{\underline{\underline{\mathfrak{Answer}}}}

y = \frac{2xz}{x+z}

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\color{orange}\huge{\underline{\underline{\mathfrak{Solution:-}}}}

{\bold{\blue{\text{Given :-}}}}

ax = by = cz and b2 = ac

Let ax = by = cz = k

Consider,  ax = k  ⇒ a = k1/x

Similarly, we get b = k1/y and c = k1/z

Now consider, b2 = ac

⇒ (k1/y)2 = (k1/x)(k1/z)

⇒ k2/y =  k1/x + 1/z

{\blue{\text{\bold{Comparing both sides}}}}

(2/y) = (1/x) + (1/z)

⇒ (2/y) = (x + z)/xz

∴ y = (2xz)/(x + z)

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Hope it helps.

Answered by DelhiQueen
41

\color{red}\huge{\underline{\underline{\mathsf{Question}}}}

if a^x=bc b^y=ca and c^z=ab prove that x/(x+1)+y/(y+1)+z/(z+1)=2 (WITHOUT LOGARITHM)

\rule{200}3

\color{orange}\huge{\underline{\underline{\mathsf{Solution}}}}

{\bold{\orange{\text{Given :-}}}}

ax = by = cz and b2 = ac

Let ax = by = cz = k

Consider,  ax = k  ⇒ a = k1/x

Similarly, we get b = k1/y and c = k1/z

Now consider, b2 = ac

⇒ (k1/y)2 = (k1/x)(k1/z)

⇒ k2/y =  k1/x + 1/z

{\green{\text{\bold{Comparing both sides}}}}

(2/y) = (1/x) + (1/z)

⇒ (2/y) = (x + z)/xz

∴ y = (2xz)/(x + z)

\rule{300}3

{\bold{\blue{\text{Hope it helps}}}}

\rule{300}3

<body bgcolor=white> <font color=black> <marquee behaviour="slide" direction="left"

❤Delhi Queen❤

<body bgcolor=white> <font color=black> <marquee behaviour="slide" direction="right"

❤Delhi Queen❤

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