If a^x = (x + y + z)^y , a^y = (x + y + z)^z and a^z = (x + y + z)^x then x + y + z (where a is not equal to zero) =

a.0

b.a^3

c.1

d.a

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You need to use logarithmic identities, such that:

`a^x = (x + y + z)^y => ln (a^x) = ln (x + y + z)^y`

`x ln a = y*ln (x + y + z)`

`a^y = (x + y + z)^z => ln (a^y) = ln (x + y + z)^z`

`y ln a = z*ln (x + y + z)`

Reasoning by analogy, yields:

`z ln a = x*ln (x + y + z)`

Adding the relations `x ln a = y*ln (x + y + z), y ln a = z*ln (x + y + z), z ln a = x*ln (x + y + z)` , yields:

`x ln a + y ln a + z ln a = y*ln (x + y + z) + z*ln (x + y + z) + x*ln (x + y + z)`

You need to factor out ln a to the left and `ln (x + y + z)` to the right, such that:

`ln a*(x + y + z) = (ln (x + y + z))(x + y + z)`

Reducing duplicate factors, yields:

`ln a = ln (x + y + z) => x + y + z = a`

**Hence, evaluating the summation `x + y + z` , under the given conditions, yields `x + y + z = a` , thus, you need to select the answer d.**

`a^x=(x+y+z)^y` (i)

`a^y=(x+y+z)^z` (ii)

`a^z=(x+y+z)^x` (iii)

multiply (i) ,(ii), and (iii) ,LHS to LHS and RHS to RHS. We have

`a^x xx a^y xxa^z=(x+y+z)^y xx(x+y+z)^z xx(x+y+z)^x`

`a^(x+y+z)=(x+y+z)^(x+y+z)`

Exponents are same on both side.

so bases are also same

a=x+y+z

ans.

corrct ans . d part

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