Dgdrxrt

Why do airplanes bank sharply to the right after air-to-air refueling?

How do we know the LHC results are robust?

How to avoid supervisors with prejudiced views?

Received an invoice from my ex-employer billing me for training; how to handle?

Preparing Indesign booklet with .psd graphics for print

Are there any unintended negative consequences to allowing PCs to gain multiple levels at once in a short milestone-XP game?

How do I transpose the 1st and -1th levels of an arbitrarily nested array?

What was the first Unix version to run on a microcomputer?

How do I make a variable always equal to the result of some calculations?

Is it professional to write unrelated content in an almost-empty email?

Won the lottery - how do I keep the money?

What benefits would be gained by using human laborers instead of drones in deep sea mining?

What is the purpose of the Evocation wizard's Potent Cantrip feature?

Return the Closest Prime Number

If the heap is initialized for security, then why is the stack uninitialized?

Make solar eclipses exceedingly rare, but still have new moons

Would this house-rule that treats advantage as a +1 to the roll instead (and disadvantage as -1) and allows them to stack be balanced?

Sending manuscript to multiple publishers

MessageLevel in QGIS3

Would a galaxy be visible from outside, but nearby?

How to start emacs in "nothing" mode (`fundamental-mode`)

What flight has the highest ratio of time difference to flight time?

Parametric curve length - calculus

Is there an analogue of projective spaces for proper schemes?

Find all real numbers satisfying the given equation.

2

$begingroup$

I came across a question which required solving two equations in real numbers $(x,y) $.
The two equations were:

beginalign
log_3x +log_2y &=2 \
3^x-2^y &= 23
endalign

Now, an obvious solution is $(3,2) $. But I want to know that how do we actually solve such equations with both exponentials and logarithms simultaneously? I tried substituting the logarithmic terms but that gave me more complicated terms in the second equation which was hard to deal with. Please help.

algebra-precalculus

share|cite|improve this question

edited 2 days ago

Brian

1,200116

asked 2 days ago

Shashwat1337Shashwat1337

10710

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Raise the first equation to the third power, and take $log_3$ of the second. You should be able to solve for $y$ with that
  $endgroup$
  – Don Thousand
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry but I still don't get it. Would you please give a little more hint?
  $endgroup$
  – Shashwat1337
  2 days ago
  

  
  
  
  

add a comment | 

2

$begingroup$

I came across a question which required solving two equations in real numbers $(x,y) $.
The two equations were:

beginalign
log_3x +log_2y &=2 \
3^x-2^y &= 23
endalign

Now, an obvious solution is $(3,2) $. But I want to know that how do we actually solve such equations with both exponentials and logarithms simultaneously? I tried substituting the logarithmic terms but that gave me more complicated terms in the second equation which was hard to deal with. Please help.

algebra-precalculus

share|cite|improve this question

edited 2 days ago

Brian

1,200116

asked 2 days ago

Shashwat1337Shashwat1337

10710

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Raise the first equation to the third power, and take $log_3$ of the second. You should be able to solve for $y$ with that
  $endgroup$
  – Don Thousand
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry but I still don't get it. Would you please give a little more hint?
  $endgroup$
  – Shashwat1337
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

I came across a question which required solving two equations in real numbers $(x,y) $.
The two equations were:

beginalign
log_3x +log_2y &=2 \
3^x-2^y &= 23
endalign

Now, an obvious solution is $(3,2) $. But I want to know that how do we actually solve such equations with both exponentials and logarithms simultaneously? I tried substituting the logarithmic terms but that gave me more complicated terms in the second equation which was hard to deal with. Please help.

algebra-precalculus

share|cite|improve this question

edited 2 days ago

Brian

1,200116

asked 2 days ago

Shashwat1337Shashwat1337

10710

$endgroup$

share|cite|improve this question

edited 2 days ago

Brian

1,200116

asked 2 days ago

Shashwat1337Shashwat1337

10710

share|cite|improve this question

edited 2 days ago

Brian

1,200116

asked 2 days ago

Shashwat1337Shashwat1337

10710

edited 2 days ago

Brian

1,200116

edited 2 days ago

Brian

1,200116

asked 2 days ago

Shashwat1337Shashwat1337

10710

asked 2 days ago

Shashwat1337Shashwat1337

10710

1 Answer
1

active

oldest

votes

4

$begingroup$

If you increase $x$, the left-hand sides of both equations increase. If you increase $y$, the left-hand side of the first equation increases, while the left-hand side of the second equation decreases.

That means that if there is some other solution $(a,b)$ and, let's say $a>3$, then the first equation says that $b<2$ while the second equation says that $b>2$.

So there are no more solutions (at least not among the positive reals).

share|cite|improve this answer

answered 2 days ago

ArthurArthur

120k7121206

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I upvoted, but you can restrict to positive reals since $log_a x$ is undefined in the reals for nonpositive reals.
  $endgroup$
  – InterstellarProbe
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much. The solution was great.
  $endgroup$
  – Shashwat1337
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @InterstellarProbe I thought I made that restriction clear enough. Do you want me to make it clearer?
  $endgroup$
  – Arthur
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur, I read the last statement as saying there might be solutions among the nonpositive reals.
  $endgroup$
  – InterstellarProbe
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur, I understand, but suppose it could be well-defined. For all $xle 0$, we have $log_3 x notin mathbbR$. So, if $xle 0, y>0$ or $x>0, yle 0$, you have $log_3 x + log_2 y notin mathbbR$. Therefore, if a solution were to exist over nonpositive real numbers, you need $x<0,y<0$ (since the logarithm cannot be defined at zero without including a point at infinity where sums are not really possible). Now, for all $x<0, y<0$, you have $0<3^x<1$ and $0<2^y<1$ which means $-1<3^x-2^y<1$, which does not put it anywhere close to 23. So, negative solutions are not possible.
  $endgroup$
  – InterstellarProbe
  2 days ago
  

  
  
  
  

 | 
show 2 more comments

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3164633%2ffind-all-real-numbers-satisfying-the-given-equation%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

If you increase $x$, the left-hand sides of both equations increase. If you increase $y$, the left-hand side of the first equation increases, while the left-hand side of the second equation decreases.

That means that if there is some other solution $(a,b)$ and, let's say $a>3$, then the first equation says that $b<2$ while the second equation says that $b>2$.

So there are no more solutions (at least not among the positive reals).

share|cite|improve this answer

answered 2 days ago

ArthurArthur

120k7121206

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I upvoted, but you can restrict to positive reals since $log_a x$ is undefined in the reals for nonpositive reals.
  $endgroup$
  – InterstellarProbe
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much. The solution was great.
  $endgroup$
  – Shashwat1337
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @InterstellarProbe I thought I made that restriction clear enough. Do you want me to make it clearer?
  $endgroup$
  – Arthur
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur, I read the last statement as saying there might be solutions among the nonpositive reals.
  $endgroup$
  – InterstellarProbe
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur, I understand, but suppose it could be well-defined. For all $xle 0$, we have $log_3 x notin mathbbR$. So, if $xle 0, y>0$ or $x>0, yle 0$, you have $log_3 x + log_2 y notin mathbbR$. Therefore, if a solution were to exist over nonpositive real numbers, you need $x<0,y<0$ (since the logarithm cannot be defined at zero without including a point at infinity where sums are not really possible). Now, for all $x<0, y<0$, you have $0<3^x<1$ and $0<2^y<1$ which means $-1<3^x-2^y<1$, which does not put it anywhere close to 23. So, negative solutions are not possible.
  $endgroup$
  – InterstellarProbe
  2 days ago
  

  
  
  
  

 | 
show 2 more comments

4

$begingroup$

If you increase $x$, the left-hand sides of both equations increase. If you increase $y$, the left-hand side of the first equation increases, while the left-hand side of the second equation decreases.

That means that if there is some other solution $(a,b)$ and, let's say $a>3$, then the first equation says that $b<2$ while the second equation says that $b>2$.

So there are no more solutions (at least not among the positive reals).

share|cite|improve this answer

answered 2 days ago

ArthurArthur

120k7121206

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I upvoted, but you can restrict to positive reals since $log_a x$ is undefined in the reals for nonpositive reals.
  $endgroup$
  – InterstellarProbe
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much. The solution was great.
  $endgroup$
  – Shashwat1337
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @InterstellarProbe I thought I made that restriction clear enough. Do you want me to make it clearer?
  $endgroup$
  – Arthur
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur, I read the last statement as saying there might be solutions among the nonpositive reals.
  $endgroup$
  – InterstellarProbe
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur, I understand, but suppose it could be well-defined. For all $xle 0$, we have $log_3 x notin mathbbR$. So, if $xle 0, y>0$ or $x>0, yle 0$, you have $log_3 x + log_2 y notin mathbbR$. Therefore, if a solution were to exist over nonpositive real numbers, you need $x<0,y<0$ (since the logarithm cannot be defined at zero without including a point at infinity where sums are not really possible). Now, for all $x<0, y<0$, you have $0<3^x<1$ and $0<2^y<1$ which means $-1<3^x-2^y<1$, which does not put it anywhere close to 23. So, negative solutions are not possible.
  $endgroup$
  – InterstellarProbe
  2 days ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

If you increase $x$, the left-hand sides of both equations increase. If you increase $y$, the left-hand side of the first equation increases, while the left-hand side of the second equation decreases.

That means that if there is some other solution $(a,b)$ and, let's say $a>3$, then the first equation says that $b<2$ while the second equation says that $b>2$.

So there are no more solutions (at least not among the positive reals).

share|cite|improve this answer

answered 2 days ago

ArthurArthur

120k7121206

$endgroup$

share|cite|improve this answer

answered 2 days ago

ArthurArthur

120k7121206

answered 2 days ago

ArthurArthur

120k7121206

answered 2 days ago

ArthurArthur

120k7121206