Question

Solve the above question

Don't use trigonometry or logarithm

Solve this with complete explanation..


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Answer 1

Answer:

Proved.

Step-by-step explanation:

Let:

$\tt{2^{x} = 3^{y} = 6^{-z} = k}$

Thus:

$\tt{2 = k^{\frac{1}{x}}}$

$\tt{3 = k^{\frac{1}{y}}}$

$\tt{6 = k^{\frac{1}{-z}}}$

We know that: 6 = 2 × 3

Therefore:

$\tt{k^{\frac{1}{x}} × k^{\frac{1}{y}} = k^{\frac{1}{-z}}}$

Base same, powers only:

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

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

hence proved.

Hope it Helps!!

Answer 2

Answer:

ACCORDING TO THE QUESTION :-

${2}^{x} = {3}^{y} = {6}^{ - z}$

TO PROVE :-

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

Suppose that -

${2}^{x} = {3}^{y} = {6}^{ - z} = p$

WE CAN WRITE THIS AS :-

$2 = {p}^{ \frac{1}{x} }$

$3 = {p}^{ \frac{1}{y} }$

$6 = {p}^{ - \frac{1}{z} }$

WE KNOW THAT -

6 = 2 × 3

=》

${p}^{ - \frac{1}{z} } = {p}^{ \frac{1}{x} } \times {p}^{ \frac{1}{y} }$

But we know the formula -

${a}^{b} \times {a}^{c} = {a}^{b + c}$

So -

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

compare -

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

HENCE PROVED....