Question

If $x=1+\log_{a}bc, \ y=1+\log_{b}ca, \ y=1+\log_{c}ab$, prove that xy + yz + zx = xyz.

Answer 1

Answer:

Step-by-step explanation:

Hi,

Given that x = 1 + logₐ(bc)

Writing 1 as logₐ(a), we can rewrite x as

x = logₐ(a) + logₐ(bc)

Using Additive Property of logarithm

log m + log n = log mn, we get

x = logₐ(abc)

By Using change of base property

x = log abc/log a

So, 1/x = log a/log abc------(1)

Similarly as done for simplifying x,

Given y = 1 + log b(ca),

so we get 1/y = log b/log abc

Given z = 1 + log c(ab)

so we get 1/z = log c/log abc

Consider

1/x + 1/y + 1/z

= log a/log abc + log b/log abc + log c/log abc

= (log a + log b + log c)/log abc

Using Additive Property of logarithm

= log abc/log abc

= 1

Hence , 1/x + 1/y + 1/z = 1

Simplifying we get

xy + yz + zx = xyz

Hence, Proved.

Hope, it helps !

Answer 2

Solution :

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We know that ,

i ) If $log_{a}x = N => a^{N} = x $

ii ) $log_{a}x + log_{a}y = log_{a}xy$

iii ) If $a^{m} = a^{n} then m = n $

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$x=1+\log_{a}bc, \ y=1+\log_{b}ca, \ y=1+\log_{c}ab$

i ) $x=1+\log_{a}bc$

=> [tex]$x = log_{a}a+\log_{a}bc$[\tex]

=> [tex]$x= log_{a}abc$[\tex]

=> $a^{x} = abc$

=>$ a = (abc)^{1/x}$ ------( 1 )

Similarly ,

$b = (abc)^{1/y}$ ---------( 2 )

$c = ( abc )^{1/z}$ ------( 3 )

Now ,

Multiply ( 1 ), ( 2 ) and ( 3 ), we get

$abc = (abc)^{1/x} (abc)^{1/y} (abc)^{1/z}$

=> $(abc)^{1} = (abc)^{1/x+1/y+1/z}$

=> 1 = 1/x + 1/y + 1/z

=> 1 = ( yz + xz + xy )/xyz

=> xyz = xy + yz + zx

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