Quite Easy One

Level pending

For three number $x$ , $y$ , $z$ , the following holds.

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

Evaluate $\boxed{\frac { 1 }{ x } +\frac { 1 }{ y } -\frac { 1 }{ z }}$

The answer is 0.

1 solution

Min-woo Lee
Apr 30, 2014

Let ${ 3 }^{ x }={ 2 }^{ y }={ 6 }^{ z }=t$ .

Then we can see that,

${ t }^{ \frac { 1 }{ x } }=3$

${ t }^{ \frac { 1 }{ y } }=2$

${ t }^{ \frac { 1 }{ z } }=6$

From law of exponent , we get

${ t }^{ \frac { 1 }{ x } +\frac { 1 }{ y } -\frac { 1 }{ z } }=\frac { 3\cdot 2 }{ 6 }=1$

$\quad \frac { 1 }{ x } +\frac { 1 }{ y } -\frac { 1 }{ z } =\log _{ t }{ \frac { 3\cdot 2 }{ 6 } } =\log _{ t }{ 1 } =0$

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

Hoe did you get that answer 3.2/6

Shabaz Rahman - 7 years, 1 month ago

Easy! It is known that for any real number a,r, and s, the following holds.

${ a }^{ r }\cdot { a }^{ s }={ a }^{ r+s }$

${ a }^{ r }\div { a }^{ s }={ a }^{ r-s }$

Therefore, since ${ t }^{ \frac { 1 }{ x } }=3$ , ${ t }^{ \frac { 1 }{ y } }=2$ , ${ t }^{ \frac { 1 }{ z } }=6$ ,

${ t }^{ \frac { 1 }{ x } }\cdot { t }^{ \frac { 1 }{ y } }= 3\cdot 2 ={ t }^{ \frac { 1 }{ x } +\frac { 1 }{ y } }$ ,

$\left( { t }^{ \frac { 1 }{ x } }\cdot { t }^{ \frac { 1 }{ y } } \right) \div { t }^{ \frac { 1 }{ z } }= \left( 3\cdot 2 \right) \div 6 ={ t }^{ \frac { 1 }{ x } +\frac { 1 }{ y } -\frac { 1 }{ z } }$ .

$\therefore \frac { 3\cdot 2 }{ 6 } ={ t }^{ \frac { 1 }{ x } +\frac { 1 }{ y } -\frac { 1 }{ z } }$

Min-woo Lee - 7 years, 1 month ago

But if we use logic, then x=y=z = 0 1/x = 1/0.

The ans should be infinity imo

Adrianus Felix - 5 years, 7 months ago

Quite Easy One

The answer is 0.

1 solution

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