Question

If 2^x=3^y=6^-z, show that 1/x + 1/y + 1/z = 0, without using Log()​

Answer 1

Solution :-

Let 2^x = 3^y = 6^-z = k

1) 2^x = k

⇒ 2 = k^(1/x)

2) 3^y = k

⇒ 3 = k^(1/y)

3) 6^-z = k

⇒ 6 = k^(-1/z)

We know that

2 * 3 = 6

⇒ k^(1/x) * k^(1/y) = k^(-1/z)

⇒ k^(1/x + 1/y) = k^(-1/z)

[ Because a^m * a^n = a^(m + n) ]

Since bases are equal we can equate powers

⇒ 1/x + 1/y = - 1/z

⇒ 1/x + 1/y + 1/z = 0

Hence shown.

Answer 2

Question :- If 2^x = 3^y = 6^(-z) then show that 1/x + 1/y + 1/z = 0 .

Solution :--

Let us assume that 2^x = 3^y = 6^(-z) = k ( where k is any constant Number ).

So, we have now,

2^x = k

→ 2 = k^(1/x),

and,

3^y = k

→ 3 = k^(1/y)

and , similarly,

6 = k^(1/(-z))

Now, we know that , 2*3 = 6

Putting all values we get,

→ [k^(1/x)] * [k^(1/y)] = k^(1/(-z))

using (a^m * a^n = a^(m+n) ) Now in LHS ,

→ k^(1/x +1/y)= k^(1/(-z))

Comparing now , we get,

→ 1/x +1/y = 1/(-z)

or,

→ 1/x + 1/y + 1/z = 0 .

✪✪ Hence Proved ✪✪

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