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)
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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
Solve without using logarithm
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
(2/y) = (1/x) + (1/z)
⇒ (2/y) = (x + z)/xz
∴ y = (2xz)/(x + z)
Hope it helps.
Answered by
41
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)
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
(2/y) = (1/x) + (1/z)
⇒ (2/y) = (x + z)/xz
∴ y = (2xz)/(x + z)
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