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Given that $log_8(x+2)$ $+$ $log_8y$ $=$ $z-frac13$ and $log_2(x-2)$ $-$ $log_2y$ = $2z+1$, Show that $x^2 = 32 ^z + 4 $

0

$begingroup$

Given that $log_8(x+2)$ $+$ $log_8y$ $=$ $z-frac13$ and $log_2(x-2)$ $-$ $log_2y$ = $2z+1$, Show that $x^2 = 32 ^z + 4 $. Any hints is appreciated. Thanks.

algebra-precalculus logarithms

share|cite|improve this question

edited Mar 31 at 21:58

J. W. Tanner

4,7871420

asked Mar 31 at 19:57

AntonioAntonio

296

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hints: use $log(a)+log(b)=log(ab)$ and $8=2^3$
  $endgroup$
  – J. W. Tanner
  Mar 31 at 20:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I can see the pattern of $log_8((x+2)*y)$ and $log_2(fracx-2y)$ but beyond that I am too dumb to solve it. Please help. @J.W.Tanner
  $endgroup$
  – Antonio
  Mar 31 at 20:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  convert from base 8 to base 2 logarithm
  $endgroup$
  – J. W. Tanner
  Mar 31 at 20:14
  

  
  
  
  

add a comment | 

0

$begingroup$

Given that $log_8(x+2)$ $+$ $log_8y$ $=$ $z-frac13$ and $log_2(x-2)$ $-$ $log_2y$ = $2z+1$, Show that $x^2 = 32 ^z + 4 $. Any hints is appreciated. Thanks.

algebra-precalculus logarithms

share|cite|improve this question

edited Mar 31 at 21:58

J. W. Tanner

4,7871420

asked Mar 31 at 19:57

AntonioAntonio

296

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hints: use $log(a)+log(b)=log(ab)$ and $8=2^3$
  $endgroup$
  – J. W. Tanner
  Mar 31 at 20:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I can see the pattern of $log_8((x+2)*y)$ and $log_2(fracx-2y)$ but beyond that I am too dumb to solve it. Please help. @J.W.Tanner
  $endgroup$
  – Antonio
  Mar 31 at 20:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  convert from base 8 to base 2 logarithm
  $endgroup$
  – J. W. Tanner
  Mar 31 at 20:14
  

  
  
  
  

add a comment | 

$begingroup$

Given that $log_8(x+2)$ $+$ $log_8y$ $=$ $z-frac13$ and $log_2(x-2)$ $-$ $log_2y$ = $2z+1$, Show that $x^2 = 32 ^z + 4 $. Any hints is appreciated. Thanks.

algebra-precalculus logarithms

share|cite|improve this question

edited Mar 31 at 21:58

J. W. Tanner

4,7871420

asked Mar 31 at 19:57

AntonioAntonio

296

$endgroup$

share|cite|improve this question

edited Mar 31 at 21:58

J. W. Tanner

4,7871420

asked Mar 31 at 19:57

AntonioAntonio

296

share|cite|improve this question

edited Mar 31 at 21:58

J. W. Tanner

4,7871420

asked Mar 31 at 19:57

AntonioAntonio

296

edited Mar 31 at 21:58

J. W. Tanner

4,7871420

edited Mar 31 at 21:58

J. W. Tanner

4,7871420

asked Mar 31 at 19:57

AntonioAntonio

296

asked Mar 31 at 19:57

AntonioAntonio

296

2 Answers
2

active

oldest

votes

2

$begingroup$

Well, $log_8(x+2) + log_8 y = log_8y(x+2)= z-frac 13$. And $log 2x-2 + log_2 y = log_2 frac x-2y=2z + 1$

So $y(x+2) = 8^z-frac 13 = frac 8^z2$

And $ frac x-2y = 2^2z+1= 2*2^2z$.

Hmmph.... Normally I'd solve for $y$ in terms of $x$ for one and substitute that value in the other and solve for $x$ but...

Just multiply the two together to get

$y(x+2)(frac x-2y) =frac 8^z2*(2*2^2z)$

$(x + 2)(x -2) = 8^z*2^2z$

$x^2 -4 = 2^3z2^2z = 2^5z = 32^z$.

$x^2 = 32^z + 4$

share|cite|improve this answer

edited Apr 1 at 9:44

J. W. Tanner

4,7871420

answered Mar 31 at 22:32

fleabloodfleablood

1

$endgroup$

add a comment | 

2

$begingroup$

Hints:

$log a+log b=log(ab)$

$log_8c=log_2c/log_2 8,$ so $3log_8 c=log_2 c$

$2^5=32$

Let me know if you need more help. Here's an answer:

Given that $log_8(x+2)$ $+$ $log_8y$ $=$ $z-frac13$ and $log_2(x-2)$ $-$ $log_2y$ = $2z+1$, we have $log_8((x+2)y)=z-frac13$ and $log_2(fracx-2y)=2z+1$, i.e., $log_2((x+2)y)=3z-1.$ Thus, $fracx-2y=2^2z+1$ and $(x+2)y=2^3z-1.$ Multiplying these equations, $x^2-4=2^5z$, so $x^2=2^5z+4=32^z+4$.

share|cite|improve this answer

answered Mar 31 at 20:17

J. W. TannerJ. W. Tanner

4,7871420

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much, the book that I am doing still hasn't shown me the change of base formula. I get it now.
  $endgroup$
  – Antonio
  Mar 31 at 20:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome. Good
  $endgroup$
  – J. W. Tanner
  Mar 31 at 21:04
  

  
  
  
  

add a comment | 

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2

$begingroup$

Well, $log_8(x+2) + log_8 y = log_8y(x+2)= z-frac 13$. And $log 2x-2 + log_2 y = log_2 frac x-2y=2z + 1$

So $y(x+2) = 8^z-frac 13 = frac 8^z2$

And $ frac x-2y = 2^2z+1= 2*2^2z$.

Hmmph.... Normally I'd solve for $y$ in terms of $x$ for one and substitute that value in the other and solve for $x$ but...

Just multiply the two together to get

$y(x+2)(frac x-2y) =frac 8^z2*(2*2^2z)$

$(x + 2)(x -2) = 8^z*2^2z$

$x^2 -4 = 2^3z2^2z = 2^5z = 32^z$.

$x^2 = 32^z + 4$

share|cite|improve this answer

edited Apr 1 at 9:44

J. W. Tanner

4,7871420

answered Mar 31 at 22:32

fleabloodfleablood

1

$endgroup$

add a comment | 

2

$begingroup$

Well, $log_8(x+2) + log_8 y = log_8y(x+2)= z-frac 13$. And $log 2x-2 + log_2 y = log_2 frac x-2y=2z + 1$

So $y(x+2) = 8^z-frac 13 = frac 8^z2$

And $ frac x-2y = 2^2z+1= 2*2^2z$.

Hmmph.... Normally I'd solve for $y$ in terms of $x$ for one and substitute that value in the other and solve for $x$ but...

Just multiply the two together to get

$y(x+2)(frac x-2y) =frac 8^z2*(2*2^2z)$

$(x + 2)(x -2) = 8^z*2^2z$

$x^2 -4 = 2^3z2^2z = 2^5z = 32^z$.

$x^2 = 32^z + 4$

share|cite|improve this answer

edited Apr 1 at 9:44

J. W. Tanner

4,7871420

answered Mar 31 at 22:32

fleabloodfleablood

1

$endgroup$

add a comment | 

$begingroup$

Well, $log_8(x+2) + log_8 y = log_8y(x+2)= z-frac 13$. And $log 2x-2 + log_2 y = log_2 frac x-2y=2z + 1$

So $y(x+2) = 8^z-frac 13 = frac 8^z2$

And $ frac x-2y = 2^2z+1= 2*2^2z$.

Hmmph.... Normally I'd solve for $y$ in terms of $x$ for one and substitute that value in the other and solve for $x$ but...

Just multiply the two together to get

$y(x+2)(frac x-2y) =frac 8^z2*(2*2^2z)$

$(x + 2)(x -2) = 8^z*2^2z$

$x^2 -4 = 2^3z2^2z = 2^5z = 32^z$.

$x^2 = 32^z + 4$

share|cite|improve this answer

edited Apr 1 at 9:44

J. W. Tanner

4,7871420

answered Mar 31 at 22:32

fleabloodfleablood

1

$endgroup$

share|cite|improve this answer

edited Apr 1 at 9:44

J. W. Tanner

4,7871420

answered Mar 31 at 22:32

fleabloodfleablood

1

edited Apr 1 at 9:44

J. W. Tanner

4,7871420

edited Apr 1 at 9:44

J. W. Tanner

4,7871420

answered Mar 31 at 22:32

fleabloodfleablood

1

answered Mar 31 at 22:32

fleabloodfleablood

1

2

$begingroup$

Hints:

$log a+log b=log(ab)$

$log_8c=log_2c/log_2 8,$ so $3log_8 c=log_2 c$

$2^5=32$

Let me know if you need more help. Here's an answer:

Given that $log_8(x+2)$ $+$ $log_8y$ $=$ $z-frac13$ and $log_2(x-2)$ $-$ $log_2y$ = $2z+1$, we have $log_8((x+2)y)=z-frac13$ and $log_2(fracx-2y)=2z+1$, i.e., $log_2((x+2)y)=3z-1.$ Thus, $fracx-2y=2^2z+1$ and $(x+2)y=2^3z-1.$ Multiplying these equations, $x^2-4=2^5z$, so $x^2=2^5z+4=32^z+4$.

share|cite|improve this answer

answered Mar 31 at 20:17

J. W. TannerJ. W. Tanner

4,7871420

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much, the book that I am doing still hasn't shown me the change of base formula. I get it now.
  $endgroup$
  – Antonio
  Mar 31 at 20:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome. Good
  $endgroup$
  – J. W. Tanner
  Mar 31 at 21:04
  

  
  
  
  

add a comment | 

2

$begingroup$

Hints:

$log a+log b=log(ab)$

$log_8c=log_2c/log_2 8,$ so $3log_8 c=log_2 c$

$2^5=32$

Let me know if you need more help. Here's an answer:

Given that $log_8(x+2)$ $+$ $log_8y$ $=$ $z-frac13$ and $log_2(x-2)$ $-$ $log_2y$ = $2z+1$, we have $log_8((x+2)y)=z-frac13$ and $log_2(fracx-2y)=2z+1$, i.e., $log_2((x+2)y)=3z-1.$ Thus, $fracx-2y=2^2z+1$ and $(x+2)y=2^3z-1.$ Multiplying these equations, $x^2-4=2^5z$, so $x^2=2^5z+4=32^z+4$.

share|cite|improve this answer

answered Mar 31 at 20:17

J. W. TannerJ. W. Tanner

4,7871420

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much, the book that I am doing still hasn't shown me the change of base formula. I get it now.
  $endgroup$
  – Antonio
  Mar 31 at 20:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome. Good
  $endgroup$
  – J. W. Tanner
  Mar 31 at 21:04
  

  
  
  
  

add a comment | 

$begingroup$

Hints:

$log a+log b=log(ab)$

$log_8c=log_2c/log_2 8,$ so $3log_8 c=log_2 c$

$2^5=32$

Let me know if you need more help. Here's an answer:

Given that $log_8(x+2)$ $+$ $log_8y$ $=$ $z-frac13$ and $log_2(x-2)$ $-$ $log_2y$ = $2z+1$, we have $log_8((x+2)y)=z-frac13$ and $log_2(fracx-2y)=2z+1$, i.e., $log_2((x+2)y)=3z-1.$ Thus, $fracx-2y=2^2z+1$ and $(x+2)y=2^3z-1.$ Multiplying these equations, $x^2-4=2^5z$, so $x^2=2^5z+4=32^z+4$.

share|cite|improve this answer

answered Mar 31 at 20:17

J. W. TannerJ. W. Tanner

4,7871420

$endgroup$

share|cite|improve this answer

answered Mar 31 at 20:17

J. W. TannerJ. W. Tanner

4,7871420

answered Mar 31 at 20:17

J. W. TannerJ. W. Tanner

4,7871420

answered Mar 31 at 20:17

J. W. TannerJ. W. Tanner

4,7871420