Question

If 2x = 3y = 6-z,then prove that 1/x + 1/y +1/z = 0​

Answer 1

Correct Question :- If 2^x = 3^y = 6^(-z) , then prove that (1/x + 1/y + 1/z) = 0 ?

Sᴏʟᴜᴛɪᴏɴ :-

Let us Assume that, 2^x=3^y=6^(-z) = k . where k is any constant Integer. (k ≠ 0) .

Than,

→ 2^x = k

→ 2 = k^(1/x)

Similarly,

→ 3^y = k

→ 3 = k^(1/y)

And,

→ 6^(-z) = k

→ 6 = k^(-1/z)

Now, since , 2 * 3 = 6 .

Putting values we get,

→ k^(1/x) * k^(1/y) = k^(-1/z)

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

→ k^(1/x + 1/y) = k^(-1/z)

Now, using , a^m = a^n => m = n

→ (1/x + 1/y) = (-1/z)

→ (1/x + 1/y + 1/z) = 0 (Proved).

Answer 2

${ \huge{ \mathfrak{ \underline{ \underline{ \pink{Question-}}}}}}$



▪ If


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


then prove that


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



${ \huge{ \mathfrak{ \underline{ \underline{ \pink{Solution-}}}}}}$



▪ Let...


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



Here, K is a constant ( K is not equal to zero)



${ \star{ \bold{ \: \: \: {2}^{x} = k}}}$



${ \implies{ \boxed{ \bold{ \red{ \: \: 2 = {k}^{ \frac{1}{x} \: \: }}}}} }$



${ \star{ \bold{ \: \: \: {3}^{y} = k}}}$



${ \implies{ \boxed{ \bold{ \red{ \: \: 3 = {k}^{ \frac{1}{y} \: \: }}}}} }$



${ \star{ \bold{ \: \: \: {6}^{ - z} = k}}}$



${ \implies{ \boxed{ \bold{ \red{ \: \: 6 = {k}^{ - \frac{1}{z} \: \: }}}}} }$



${ \bold{ \underline{ \pink{ since\: \: \: 2 \times 3 = 6 \: \: \: \: \: \: \: }}}}$



▪ substituting the values of 2, 3 and 6 derived from the given question....



${ \boxed{ \bold{ \purple{ \: \: {k}^{ \frac{1}{x} } \times {k}^{ \frac{1}{y} } = {k}^{ - \frac{1}{z}} \: \: \: \: }}} }$



we know that,



${ \bold{ \underline{ \pink{ \: {a}^{b} \times {a}^{c} = {a}^{(b + c) \: }}}}}$



using this identity in L.H.S....



${ \bold{ \implies{ \purple{ {k}^{( \frac{1}{x} + \frac{1}{y} )} = {k}^{ - \frac{1}{z}}}}} }$



${ \bold{ \implies{ \purple{ \frac{1}{x} + \frac{1}{y} = - \frac{1}{z}}}}}$



${ \boxed{ \bold{ \implies{ \red{ \: \: \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \: \: }}}}}$



hence, proved....