Math, asked by Prabal5499, 10 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

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

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

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

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

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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

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

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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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