Question

if a^1/x = b^1/y = c^1/z and abc=1 and prove that x+ y +z=0

please explain the question

Answer 1

Hi ,

Here is your answer !
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Given that,

${a}^{ \frac{1}{x} } = {b}^{ \frac{1}{y} } = {c}^{ \frac{1}{z} }$
Let
${a}^{ \frac{1}{x} } = {b}^{ \frac{1}{y} } = {c}^{ \frac{1}{z} } = k \\ \\ \implies \:a = {k}^{x} \:, \: b= {k}^{y} \:, \: c = {k}^{z}$

Now ,

Also given that

$abc = 1 \\ \\ \implies {k}^{x} \times {k}^{y} \times {k}^{z} = 1 \\ \\ \implies {k}^{x + y + z} = 1 \\ \\ \implies{k}^{x + y + z} = {k}^{0} \: \: \: \: \: \: \: ( \: \because {k}^{0} = 1 \: ) \\ \\ on \: \: equating \: \: power \: \: of \: \: k \\ we \: get\\ \\ {x + y + z} =0 \: \: \: \: \: [\:Proved\:]$

Answer 2

Question :- if a^(1/x) = b^(1/y) = c^(1/z) , and abc = 1 , prove that, x + y + z = 0 .

Solution :-

Let us assume that,

→ a^(1/x) = b^(1/y) = c^(1/z) = k , (where k is a constant number.)

then,

→ a^(1/x) = k

→ a = k^x

similarly,

→ b^(1/y) = k

→ b = k^y

and,

→ c^(1/z) = k

→ c = k^z .

given that,

→ a * b * c = 1 .

putting above values we get,

→ k^x * k^y * k^z = 1

using a^l * a^m * a^n = a^(l + m + n) , we get,

→ k^(x + y + z) = 1

now, we know that, if power of a number is 0 , it is equal to 1,

therefore,

→ k^(x + y + z) = k^0

now, we know that, when base is same, powers are equal . { a^m = a^n => m = n }

hence,

→ (x + y + z) = 0 (Proved.)

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