Mixing
Compute $2^{2x}-2^{-2x}$
$begingroup$
$$2^x - 2^{-x} = 4$$
- Compute $2^{2x}-2^{-2x}$
I'm unable to notice anything here. Maybe we can square the first equation in order to get that expression. Could you assist me with this?
Regards
algebra-precalculus
asked Jan 10 at 13:43
EnzoEnzo
1687
$endgroup$
add a comment |
$begingroup$
$$2^x - 2^{-x} = 4$$
- Compute $2^{2x}-2^{-2x}$
I'm unable to notice anything here. Maybe we can square the first equation in order to get that expression. Could you assist me with this?
Regards
algebra-precalculus
asked Jan 10 at 13:43
EnzoEnzo
1687
$endgroup$
$begingroup$
Well, what do you get if you square both sides?
$endgroup$
– lulu
Jan 10 at 13:46
$begingroup$
Subsitute $2^{x}$ by t>0; is it any easier now?
$endgroup$
– The Cat
Jan 10 at 13:49
add a comment |
$begingroup$
$$2^x - 2^{-x} = 4$$
- Compute $2^{2x}-2^{-2x}$
I'm unable to notice anything here. Maybe we can square the first equation in order to get that expression. Could you assist me with this?
Regards
algebra-precalculus
asked Jan 10 at 13:43
EnzoEnzo
1687
$endgroup$
$$2^x - 2^{-x} = 4$$
- Compute $2^{2x}-2^{-2x}$
I'm unable to notice anything here. Maybe we can square the first equation in order to get that expression. Could you assist me with this?
Regards
algebra-precalculus
algebra-precalculus
asked Jan 10 at 13:43
EnzoEnzo
1687
asked Jan 10 at 13:43
EnzoEnzo
1687
asked Jan 10 at 13:43
EnzoEnzo
1687
asked Jan 10 at 13:43
EnzoEnzo
1687
asked Jan 10 at 13:43
EnzoEnzo
1687
1687
$begingroup$
Well, what do you get if you square both sides?
$endgroup$
– lulu
Jan 10 at 13:46
$begingroup$
Subsitute $2^{x}$ by t>0; is it any easier now?
$endgroup$
– The Cat
Jan 10 at 13:49
add a comment |
$begingroup$
Well, what do you get if you square both sides?
$endgroup$
– lulu
Jan 10 at 13:46
$begingroup$
Subsitute $2^{x}$ by t>0; is it any easier now?
$endgroup$
– The Cat
Jan 10 at 13:49
$begingroup$
Well, what do you get if you square both sides?
$endgroup$
– lulu
Jan 10 at 13:46
$begingroup$
Well, what do you get if you square both sides?
$endgroup$
– lulu
Jan 10 at 13:46
$begingroup$
Subsitute $2^{x}$ by t>0; is it any easier now?
$endgroup$
– The Cat
Jan 10 at 13:49
$begingroup$
Subsitute $2^{x}$ by t>0; is it any easier now?
$endgroup$
– The Cat
Jan 10 at 13:49
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: Let $t = 2^x$ so that $2^{-x} = t^{-1} = frac{1}{t}$. Hence, the equation simplifies to
$$t-frac{1}{t} = 4$$
You’re left with the equation above. Can you continue from here? Note that a non-positive solution is extraneous because $2^x > 0$. Finally, you would have to evaluate $2^{2x}-2^{-2x}$, which is
$$t^2-frac{1}{t^2}$$
The key here is to realize that such questions typically give extra information to throw you off, so always look for good substitutions to simplify these problems.
answered Jan 10 at 13:49
KM101KM101
5,9251524
$endgroup$
add a comment |
$begingroup$
We have $a-dfrac1a=4$
We need $a^2-dfrac1{a^2}=left(a-dfrac1aright)left(a+dfrac1aright)$
$$left(a+dfrac1aright)^2=left(a-dfrac1aright)^2+4acdotdfrac1a=?$$
answered Jan 10 at 13:47
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
add a comment |
$begingroup$
$$2^x-2^{-x}=4$$
$$2^{2x}-1=4times2^x$$
$$(2^x)^2-4(2^x)-1=0$$
let $u=2^x$ and now solve:
$$u^2-4u-1=0$$
answered Jan 10 at 14:32
Henry LeeHenry Lee
1,894219
$endgroup$
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
});
}
});
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%2f3068655%2fcompute-22x-2-2x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Let $t = 2^x$ so that $2^{-x} = t^{-1} = frac{1}{t}$. Hence, the equation simplifies to
$$t-frac{1}{t} = 4$$
You’re left with the equation above. Can you continue from here? Note that a non-positive solution is extraneous because $2^x > 0$. Finally, you would have to evaluate $2^{2x}-2^{-2x}$, which is
$$t^2-frac{1}{t^2}$$
The key here is to realize that such questions typically give extra information to throw you off, so always look for good substitutions to simplify these problems.
answered Jan 10 at 13:49
KM101KM101
5,9251524
$endgroup$
add a comment |
$begingroup$
Hint: Let $t = 2^x$ so that $2^{-x} = t^{-1} = frac{1}{t}$. Hence, the equation simplifies to
$$t-frac{1}{t} = 4$$
You’re left with the equation above. Can you continue from here? Note that a non-positive solution is extraneous because $2^x > 0$. Finally, you would have to evaluate $2^{2x}-2^{-2x}$, which is
$$t^2-frac{1}{t^2}$$
The key here is to realize that such questions typically give extra information to throw you off, so always look for good substitutions to simplify these problems.
answered Jan 10 at 13:49
KM101KM101
5,9251524
$endgroup$
add a comment |
$begingroup$
Hint: Let $t = 2^x$ so that $2^{-x} = t^{-1} = frac{1}{t}$. Hence, the equation simplifies to
$$t-frac{1}{t} = 4$$
You’re left with the equation above. Can you continue from here? Note that a non-positive solution is extraneous because $2^x > 0$. Finally, you would have to evaluate $2^{2x}-2^{-2x}$, which is
$$t^2-frac{1}{t^2}$$
The key here is to realize that such questions typically give extra information to throw you off, so always look for good substitutions to simplify these problems.
answered Jan 10 at 13:49
KM101KM101
5,9251524
$endgroup$
Hint: Let $t = 2^x$ so that $2^{-x} = t^{-1} = frac{1}{t}$. Hence, the equation simplifies to
$$t-frac{1}{t} = 4$$
You’re left with the equation above. Can you continue from here? Note that a non-positive solution is extraneous because $2^x > 0$. Finally, you would have to evaluate $2^{2x}-2^{-2x}$, which is
$$t^2-frac{1}{t^2}$$
The key here is to realize that such questions typically give extra information to throw you off, so always look for good substitutions to simplify these problems.
answered Jan 10 at 13:49
KM101KM101
5,9251524
edited Jan 10 at 14:00
answered Jan 10 at 13:49
KM101KM101
5,9251524
answered Jan 10 at 13:49
KM101KM101
5,9251524
answered Jan 10 at 13:49
KM101KM101
5,9251524
5,9251524
add a comment |
add a comment |
$begingroup$
We have $a-dfrac1a=4$
We need $a^2-dfrac1{a^2}=left(a-dfrac1aright)left(a+dfrac1aright)$
$$left(a+dfrac1aright)^2=left(a-dfrac1aright)^2+4acdotdfrac1a=?$$
answered Jan 10 at 13:47
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
add a comment |
$begingroup$
We have $a-dfrac1a=4$
We need $a^2-dfrac1{a^2}=left(a-dfrac1aright)left(a+dfrac1aright)$
$$left(a+dfrac1aright)^2=left(a-dfrac1aright)^2+4acdotdfrac1a=?$$
answered Jan 10 at 13:47
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
add a comment |
$begingroup$
We have $a-dfrac1a=4$
We need $a^2-dfrac1{a^2}=left(a-dfrac1aright)left(a+dfrac1aright)$
$$left(a+dfrac1aright)^2=left(a-dfrac1aright)^2+4acdotdfrac1a=?$$
answered Jan 10 at 13:47
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
We have $a-dfrac1a=4$
We need $a^2-dfrac1{a^2}=left(a-dfrac1aright)left(a+dfrac1aright)$
$$left(a+dfrac1aright)^2=left(a-dfrac1aright)^2+4acdotdfrac1a=?$$
answered Jan 10 at 13:47
lab bhattacharjeelab bhattacharjee
225k15157275
answered Jan 10 at 13:47
lab bhattacharjeelab bhattacharjee
225k15157275
answered Jan 10 at 13:47
lab bhattacharjeelab bhattacharjee
225k15157275
answered Jan 10 at 13:47
lab bhattacharjeelab bhattacharjee
225k15157275
225k15157275
add a comment |
add a comment |
$begingroup$
$$2^x-2^{-x}=4$$
$$2^{2x}-1=4times2^x$$
$$(2^x)^2-4(2^x)-1=0$$
let $u=2^x$ and now solve:
$$u^2-4u-1=0$$
answered Jan 10 at 14:32
Henry LeeHenry Lee
1,894219
$endgroup$
add a comment |
$begingroup$
$$2^x-2^{-x}=4$$
$$2^{2x}-1=4times2^x$$
$$(2^x)^2-4(2^x)-1=0$$
let $u=2^x$ and now solve:
$$u^2-4u-1=0$$
answered Jan 10 at 14:32
Henry LeeHenry Lee
1,894219
$endgroup$
add a comment |
$begingroup$
$$2^x-2^{-x}=4$$
$$2^{2x}-1=4times2^x$$
$$(2^x)^2-4(2^x)-1=0$$
let $u=2^x$ and now solve:
$$u^2-4u-1=0$$
answered Jan 10 at 14:32
Henry LeeHenry Lee
1,894219
$endgroup$
$$2^x-2^{-x}=4$$
$$2^{2x}-1=4times2^x$$
$$(2^x)^2-4(2^x)-1=0$$
let $u=2^x$ and now solve:
$$u^2-4u-1=0$$
answered Jan 10 at 14:32
Henry LeeHenry Lee
1,894219
answered Jan 10 at 14:32
Henry LeeHenry Lee
1,894219
answered Jan 10 at 14:32
Henry LeeHenry Lee
1,894219
answered Jan 10 at 14:32
Henry LeeHenry Lee
1,894219
1,894219
add a comment |
add a comment |
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.
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%2f3068655%2fcompute-22x-2-2x%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
$begingroup$
Well, what do you get if you square both sides?
$endgroup$
– lulu
Jan 10 at 13:46
$begingroup$
Subsitute $2^{x}$ by t>0; is it any easier now?
$endgroup$
– The Cat
Jan 10 at 13:49