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$k^x(z - y) + k^y(x - z) + k^z(y - x) neq 0$

0

$begingroup$

Let $k > 0, k neq 1$, and $x,y,z$ distinct real numbers. Show that
$$k^x(z - y) + k^y(x - z) +k^z(y - x) neq 0.$$

My progress: Assume the expression is equal to 0. Write this as
$$(z - y)k^x + k^z(y - x) = k^y(z-x).$$

Divide by $k^y$ and obtain
$$(z-y)k^x-y + (y-x)k^z - y = z - x.$$
If we set $a = x - y, b = z - y$ we get that $a,b neq 0$ and we can write the expression as
$$b k^a - ak^b = b - a iff b ( k^a - 1) = a(k^b - 1) iff frack^a - 1k^b - 1 = fracab.$$

At this point I'm stuck. Any ideas?

algebra-precalculus

share|cite|improve this question

asked Mar 19 at 21:22

Davidmath7Davidmath7

1665

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $$b k^a - ak^b = b - a $$ certainly is true if $(k=1)$ or $(a=b)$ or $(a=0)$ or $(b=0)$
  $endgroup$
  – WW1
  Mar 19 at 21:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, but the conditions rule out all of those cases.
  $endgroup$
  – Davidmath7
  Mar 19 at 22:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's exactly the point !
  $endgroup$
  – WW1
  Mar 19 at 22:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you can show that there are no other solutions, you will have proved the statement.
  $endgroup$
  – WW1
  Mar 19 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well of course, my question was pretty much how can i show that there are no other solutions.
  $endgroup$
  – Davidmath7
  Mar 19 at 22:32
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $k > 0, k neq 1$, and $x,y,z$ distinct real numbers. Show that
$$k^x(z - y) + k^y(x - z) +k^z(y - x) neq 0.$$

My progress: Assume the expression is equal to 0. Write this as
$$(z - y)k^x + k^z(y - x) = k^y(z-x).$$

Divide by $k^y$ and obtain
$$(z-y)k^x-y + (y-x)k^z - y = z - x.$$
If we set $a = x - y, b = z - y$ we get that $a,b neq 0$ and we can write the expression as
$$b k^a - ak^b = b - a iff b ( k^a - 1) = a(k^b - 1) iff frack^a - 1k^b - 1 = fracab.$$

At this point I'm stuck. Any ideas?

algebra-precalculus

share|cite|improve this question

asked Mar 19 at 21:22

Davidmath7Davidmath7

1665

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $$b k^a - ak^b = b - a $$ certainly is true if $(k=1)$ or $(a=b)$ or $(a=0)$ or $(b=0)$
  $endgroup$
  – WW1
  Mar 19 at 21:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, but the conditions rule out all of those cases.
  $endgroup$
  – Davidmath7
  Mar 19 at 22:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's exactly the point !
  $endgroup$
  – WW1
  Mar 19 at 22:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you can show that there are no other solutions, you will have proved the statement.
  $endgroup$
  – WW1
  Mar 19 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well of course, my question was pretty much how can i show that there are no other solutions.
  $endgroup$
  – Davidmath7
  Mar 19 at 22:32
  

  
  
  
  

add a comment | 

$begingroup$

Let $k > 0, k neq 1$, and $x,y,z$ distinct real numbers. Show that
$$k^x(z - y) + k^y(x - z) +k^z(y - x) neq 0.$$

My progress: Assume the expression is equal to 0. Write this as
$$(z - y)k^x + k^z(y - x) = k^y(z-x).$$

Divide by $k^y$ and obtain
$$(z-y)k^x-y + (y-x)k^z - y = z - x.$$
If we set $a = x - y, b = z - y$ we get that $a,b neq 0$ and we can write the expression as
$$b k^a - ak^b = b - a iff b ( k^a - 1) = a(k^b - 1) iff frack^a - 1k^b - 1 = fracab.$$

At this point I'm stuck. Any ideas?

algebra-precalculus

share|cite|improve this question

asked Mar 19 at 21:22

Davidmath7Davidmath7

1665

$endgroup$

share|cite|improve this question

asked Mar 19 at 21:22

Davidmath7Davidmath7

1665

share|cite|improve this question

asked Mar 19 at 21:22

Davidmath7Davidmath7

1665

asked Mar 19 at 21:22

Davidmath7Davidmath7

1665

asked Mar 19 at 21:22

Davidmath7Davidmath7

1665