Mixing
Solution of a typical equation with surds power
1
$begingroup$
I was attempting to find a solution for the equation
$1 + 12^sqrt{x} = 9^sqrt{x} + 10^sqrt{x}$. By the trial and error, I found a solution $ x = 9$. Is there any method to solve these equation?
exponential-function radicals
edited Jan 2 at 12:56
Harry Peter
5,47111439
asked Dec 17 '18 at 17:09
user144660user144660
655
$endgroup$
|
show 3 more comments
1
$begingroup$
I was attempting to find a solution for the equation
$1 + 12^sqrt{x} = 9^sqrt{x} + 10^sqrt{x}$. By the trial and error, I found a solution $ x = 9$. Is there any method to solve these equation?
exponential-function radicals
edited Jan 2 at 12:56
Harry Peter
5,47111439
asked Dec 17 '18 at 17:09
user144660user144660
655
$endgroup$
$begingroup$
Let $y=sqrt{x}$, $ygeq0$.
$endgroup$
– Don Thousand
Dec 17 '18 at 17:11
$begingroup$
I tried with this transformation. But did not work.
$endgroup$
– user144660
Dec 17 '18 at 17:12
1
$begingroup$
What do you mean by "it did not work"? It dramatically simplifies the problem.
$endgroup$
– Don Thousand
Dec 17 '18 at 17:13
$begingroup$
By 'did not work', I meant this transformation did not solve the problem. DO you get the solution AT ALL by this method?
$endgroup$
– user144660
Dec 17 '18 at 17:14
2
$begingroup$
I mean, it's easy to show that no solution exists with $y>3$, and via intermediate value theorem, you can prove that there are at most two solutions to this equation. I won't do the rest, but once you find the range in which a second solution might exist, the problem will become easier.
$endgroup$
– Don Thousand
Dec 17 '18 at 17:16
|
show 3 more comments
1
1
1
$begingroup$
I was attempting to find a solution for the equation
$1 + 12^sqrt{x} = 9^sqrt{x} + 10^sqrt{x}$. By the trial and error, I found a solution $ x = 9$. Is there any method to solve these equation?
exponential-function radicals
edited Jan 2 at 12:56
Harry Peter
5,47111439
asked Dec 17 '18 at 17:09
user144660user144660
655
$endgroup$
I was attempting to find a solution for the equation
$1 + 12^sqrt{x} = 9^sqrt{x} + 10^sqrt{x}$. By the trial and error, I found a solution $ x = 9$. Is there any method to solve these equation?
exponential-function radicals
exponential-function radicals
edited Jan 2 at 12:56
Harry Peter
5,47111439
asked Dec 17 '18 at 17:09
user144660user144660
655
edited Jan 2 at 12:56
Harry Peter
5,47111439
asked Dec 17 '18 at 17:09
user144660user144660
655
edited Jan 2 at 12:56
Harry Peter
5,47111439
edited Jan 2 at 12:56
Harry Peter
5,47111439
edited Jan 2 at 12:56
Harry Peter
5,47111439
5,47111439
asked Dec 17 '18 at 17:09
user144660user144660
655
asked Dec 17 '18 at 17:09
user144660user144660
655
asked Dec 17 '18 at 17:09
user144660user144660
655
655
$begingroup$
Let $y=sqrt{x}$, $ygeq0$.
$endgroup$
– Don Thousand
Dec 17 '18 at 17:11
$begingroup$
I tried with this transformation. But did not work.
$endgroup$
– user144660
Dec 17 '18 at 17:12
1
$begingroup$
What do you mean by "it did not work"? It dramatically simplifies the problem.
$endgroup$
– Don Thousand
Dec 17 '18 at 17:13
$begingroup$
By 'did not work', I meant this transformation did not solve the problem. DO you get the solution AT ALL by this method?
$endgroup$
– user144660
Dec 17 '18 at 17:14
2
$begingroup$
I mean, it's easy to show that no solution exists with $y>3$, and via intermediate value theorem, you can prove that there are at most two solutions to this equation. I won't do the rest, but once you find the range in which a second solution might exist, the problem will become easier.
$endgroup$
– Don Thousand
Dec 17 '18 at 17:16
|
show 3 more comments
$begingroup$
Let $y=sqrt{x}$, $ygeq0$.
$endgroup$
– Don Thousand
Dec 17 '18 at 17:11
$begingroup$
I tried with this transformation. But did not work.
$endgroup$
– user144660
Dec 17 '18 at 17:12
1
$begingroup$
What do you mean by "it did not work"? It dramatically simplifies the problem.
$endgroup$
– Don Thousand
Dec 17 '18 at 17:13
$begingroup$
By 'did not work', I meant this transformation did not solve the problem. DO you get the solution AT ALL by this method?
$endgroup$
– user144660
Dec 17 '18 at 17:14
2
$begingroup$
I mean, it's easy to show that no solution exists with $y>3$, and via intermediate value theorem, you can prove that there are at most two solutions to this equation. I won't do the rest, but once you find the range in which a second solution might exist, the problem will become easier.
$endgroup$
– Don Thousand
Dec 17 '18 at 17:16
$begingroup$
Let $y=sqrt{x}$, $ygeq0$.
$endgroup$
– Don Thousand
Dec 17 '18 at 17:11
$begingroup$
Let $y=sqrt{x}$, $ygeq0$.
$endgroup$
– Don Thousand
Dec 17 '18 at 17:11
$begingroup$
I tried with this transformation. But did not work.
$endgroup$
– user144660
Dec 17 '18 at 17:12
$begingroup$
I tried with this transformation. But did not work.
$endgroup$
– user144660
Dec 17 '18 at 17:12
1
1
$begingroup$
What do you mean by "it did not work"? It dramatically simplifies the problem.
$endgroup$
– Don Thousand
Dec 17 '18 at 17:13
$begingroup$
What do you mean by "it did not work"? It dramatically simplifies the problem.
$endgroup$
– Don Thousand
Dec 17 '18 at 17:13
$begingroup$
By 'did not work', I meant this transformation did not solve the problem. DO you get the solution AT ALL by this method?
$endgroup$
– user144660
Dec 17 '18 at 17:14
$begingroup$
By 'did not work', I meant this transformation did not solve the problem. DO you get the solution AT ALL by this method?
$endgroup$
– user144660
Dec 17 '18 at 17:14
2
2
$begingroup$
I mean, it's easy to show that no solution exists with $y>3$, and via intermediate value theorem, you can prove that there are at most two solutions to this equation. I won't do the rest, but once you find the range in which a second solution might exist, the problem will become easier.
$endgroup$
– Don Thousand
Dec 17 '18 at 17:16
$begingroup$
I mean, it's easy to show that no solution exists with $y>3$, and via intermediate value theorem, you can prove that there are at most two solutions to this equation. I won't do the rest, but once you find the range in which a second solution might exist, the problem will become easier.
$endgroup$
– Don Thousand
Dec 17 '18 at 17:16
|
show 3 more comments
1 Answer
1
active
oldest
votes
0
$begingroup$
Step 1. Set $,y=sqrt{x}$, and obtain the equation
$$
f(y)=12^y-10^y-9^y+1=0
$$
Step 2. Show that $f(y)<0$, if $yin(0,3)$, $f(y)>0$, if $y>3$ or $y<0$.
answered Jan 2 at 13:39
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.9k1384163
$endgroup$
$begingroup$
Thanks a lot! I got it now.
$endgroup$
– user144660
Jan 3 at 14:10
add a comment |
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044184%2fsolution-of-a-typical-equation-with-surds-power%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
$begingroup$
Step 1. Set $,y=sqrt{x}$, and obtain the equation
$$
f(y)=12^y-10^y-9^y+1=0
$$
Step 2. Show that $f(y)<0$, if $yin(0,3)$, $f(y)>0$, if $y>3$ or $y<0$.
answered Jan 2 at 13:39
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.9k1384163
$endgroup$
$begingroup$
Thanks a lot! I got it now.
$endgroup$
– user144660
Jan 3 at 14:10
add a comment |
0
$begingroup$
Step 1. Set $,y=sqrt{x}$, and obtain the equation
$$
f(y)=12^y-10^y-9^y+1=0
$$
Step 2. Show that $f(y)<0$, if $yin(0,3)$, $f(y)>0$, if $y>3$ or $y<0$.
answered Jan 2 at 13:39
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.9k1384163
$endgroup$
$begingroup$
Thanks a lot! I got it now.
$endgroup$
– user144660
Jan 3 at 14:10
add a comment |
0
0
0
$begingroup$
Step 1. Set $,y=sqrt{x}$, and obtain the equation
$$
f(y)=12^y-10^y-9^y+1=0
$$
Step 2. Show that $f(y)<0$, if $yin(0,3)$, $f(y)>0$, if $y>3$ or $y<0$.
answered Jan 2 at 13:39
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.9k1384163
$endgroup$
Step 1. Set $,y=sqrt{x}$, and obtain the equation
$$
f(y)=12^y-10^y-9^y+1=0
$$
Step 2. Show that $f(y)<0$, if $yin(0,3)$, $f(y)>0$, if $y>3$ or $y<0$.
answered Jan 2 at 13:39
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.9k1384163
answered Jan 2 at 13:39
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.9k1384163
answered Jan 2 at 13:39
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.9k1384163
answered Jan 2 at 13:39
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.9k1384163
62.9k1384163
$begingroup$
Thanks a lot! I got it now.
$endgroup$
– user144660
Jan 3 at 14:10
add a comment |
$begingroup$
Thanks a lot! I got it now.
$endgroup$
– user144660
Jan 3 at 14:10
$begingroup$
Thanks a lot! I got it now.
$endgroup$
– user144660
Jan 3 at 14:10
$begingroup$
Thanks a lot! I got it now.
$endgroup$
– user144660
Jan 3 at 14:10
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044184%2fsolution-of-a-typical-equation-with-surds-power%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Let $y=sqrt{x}$, $ygeq0$.
$endgroup$
– Don Thousand
Dec 17 '18 at 17:11
$begingroup$
I tried with this transformation. But did not work.
$endgroup$
– user144660
Dec 17 '18 at 17:12
1
$begingroup$
What do you mean by "it did not work"? It dramatically simplifies the problem.
$endgroup$
– Don Thousand
Dec 17 '18 at 17:13
$begingroup$
By 'did not work', I meant this transformation did not solve the problem. DO you get the solution AT ALL by this method?
$endgroup$
– user144660
Dec 17 '18 at 17:14
2
$begingroup$
I mean, it's easy to show that no solution exists with $y>3$, and via intermediate value theorem, you can prove that there are at most two solutions to this equation. I won't do the rest, but once you find the range in which a second solution might exist, the problem will become easier.
$endgroup$
– Don Thousand
Dec 17 '18 at 17:16