Question

2^x=3^y=6^-z then prove that 1/x+1/y+1/z=0

Answer 1

Step-by-step explanation:

Refer to the attachment.

Answer 2

Hey mate !! Here you go :-

Let $2^{x}$ = $3^{y}$ = $6^{-z}$ = k ( constant )

From here we get,

As $2^{2}$ = 4

2 = $\sqrt{4}$

2 = $4^{1/2}$

$2^{x}$ = k

2 = $k^{1/x}$

And

$3^{y}$ = k

3 = $k^{1/y}$

Also,

$6^{-z}$ = k

6 = $k^{-1/z}$

As we know that,

3*2=6

Not put the values of 3,2 and 6

$k^{1/y}$ * $k^{1/x}$ = $k^{-1/z}$

By the law of exponent,

$a^{m}$ * $a^{n}$ = $a^{m+n}$

We get,

$k^{1/x+1/y}$ = $k^{-1/z}$

Since bases are same we can compare the powers,

→ $\frac{1}{x}$ + $\frac{1}{y}$ = $\frac{-1}{z}$

Solving this we get,

$\frac{1}{x}$ + $\frac{1}{y}$ + $\frac{1}{z}$ = 0

∴ Proved, Question solved !! :-D

Hope that helps you..

Please mark as brainliest answer..

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~