Solve this : $ x^{frac 34 { log_2^2 x} + log_2 {(x-5/4) }} = sqrt 2 $
$begingroup$
The question : $$ x^{frac 34 { log_2^2 x} + log_2 {left(x-frac 54right) }} = sqrt 2 $$
My try at it : I think that the whole equation should be converted to log with base 2. What to do next?
logarithms
asked Apr 14 '17 at 14:33
Esha MukhopadhyayEsha Mukhopadhyay
315
$endgroup$
|
show 9 more comments
$begingroup$
The question : $$ x^{frac 34 { log_2^2 x} + log_2 {left(x-frac 54right) }} = sqrt 2 $$
My try at it : I think that the whole equation should be converted to log with base 2. What to do next?
logarithms
asked Apr 14 '17 at 14:33
Esha MukhopadhyayEsha Mukhopadhyay
315
$endgroup$
$begingroup$
Does $log^2 x$ mean $log(log x)$ or $(log x)^2$
$endgroup$
– DHMO
Apr 14 '17 at 14:39
1
$begingroup$
The title and the question have different equations in them. Which one do you mean?
$endgroup$
– AsafHaas
Apr 14 '17 at 14:40
$begingroup$
$ (log_2 x) ^ 2 $
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
@AsafHaas its the same. i hav copied it from the title.
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
By executing your idea and let $y=log_2 x$, we get $1/2=y(3y^2/4+log_2{(x-5/4)})$
$endgroup$
– didgogns
Apr 14 '17 at 14:43
|
show 9 more comments
$begingroup$
The question : $$ x^{frac 34 { log_2^2 x} + log_2 {left(x-frac 54right) }} = sqrt 2 $$
My try at it : I think that the whole equation should be converted to log with base 2. What to do next?
logarithms
asked Apr 14 '17 at 14:33
Esha MukhopadhyayEsha Mukhopadhyay
315
$endgroup$
The question : $$ x^{frac 34 { log_2^2 x} + log_2 {left(x-frac 54right) }} = sqrt 2 $$
My try at it : I think that the whole equation should be converted to log with base 2. What to do next?
logarithms
logarithms
asked Apr 14 '17 at 14:33
Esha MukhopadhyayEsha Mukhopadhyay
315
asked Apr 14 '17 at 14:33
Esha MukhopadhyayEsha Mukhopadhyay
315
edited Apr 14 '17 at 15:09
Esha Mukhopadhyay
asked Apr 14 '17 at 14:33
Esha MukhopadhyayEsha Mukhopadhyay
315
asked Apr 14 '17 at 14:33
Esha MukhopadhyayEsha Mukhopadhyay
315
asked Apr 14 '17 at 14:33
Esha MukhopadhyayEsha Mukhopadhyay
315
315
$begingroup$
Does $log^2 x$ mean $log(log x)$ or $(log x)^2$
$endgroup$
– DHMO
Apr 14 '17 at 14:39
1
$begingroup$
The title and the question have different equations in them. Which one do you mean?
$endgroup$
– AsafHaas
Apr 14 '17 at 14:40
$begingroup$
$ (log_2 x) ^ 2 $
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
@AsafHaas its the same. i hav copied it from the title.
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
By executing your idea and let $y=log_2 x$, we get $1/2=y(3y^2/4+log_2{(x-5/4)})$
$endgroup$
– didgogns
Apr 14 '17 at 14:43
|
show 9 more comments
$begingroup$
Does $log^2 x$ mean $log(log x)$ or $(log x)^2$
$endgroup$
– DHMO
Apr 14 '17 at 14:39
1
$begingroup$
The title and the question have different equations in them. Which one do you mean?
$endgroup$
– AsafHaas
Apr 14 '17 at 14:40
$begingroup$
$ (log_2 x) ^ 2 $
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
@AsafHaas its the same. i hav copied it from the title.
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
By executing your idea and let $y=log_2 x$, we get $1/2=y(3y^2/4+log_2{(x-5/4)})$
$endgroup$
– didgogns
Apr 14 '17 at 14:43
$begingroup$
Does $log^2 x$ mean $log(log x)$ or $(log x)^2$
$endgroup$
– DHMO
Apr 14 '17 at 14:39
$begingroup$
Does $log^2 x$ mean $log(log x)$ or $(log x)^2$
$endgroup$
– DHMO
Apr 14 '17 at 14:39
1
1
$begingroup$
The title and the question have different equations in them. Which one do you mean?
$endgroup$
– AsafHaas
Apr 14 '17 at 14:40
$begingroup$
The title and the question have different equations in them. Which one do you mean?
$endgroup$
– AsafHaas
Apr 14 '17 at 14:40
$begingroup$
$ (log_2 x) ^ 2 $
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
$ (log_2 x) ^ 2 $
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
@AsafHaas its the same. i hav copied it from the title.
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
@AsafHaas its the same. i hav copied it from the title.
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
By executing your idea and let $y=log_2 x$, we get $1/2=y(3y^2/4+log_2{(x-5/4)})$
$endgroup$
– didgogns
Apr 14 '17 at 14:43
$begingroup$
By executing your idea and let $y=log_2 x$, we get $1/2=y(3y^2/4+log_2{(x-5/4)})$
$endgroup$
– didgogns
Apr 14 '17 at 14:43
|
show 9 more comments
1 Answer
1
active
oldest
votes
$begingroup$
taking the logarithm on both sides of your equation we obatin
$$left(frac{3}{4}left(frac{ln(x)}{ln(2)}right)^2+frac{lnleft(x-frac{5}{4}right)}{ln(2)}right)ln(x)=frac{1}{2}ln(2)$$ and here you can take a numerical method
with such a method we obtain $$xapprox 2.050105808$$
answered Apr 14 '17 at 14:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
$endgroup$
$begingroup$
this is turning out to be a tricky thing. can u please solve it?
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:58
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%2f2233993%2fsolve-this-x-frac-34-log-22-x-log-2-x-5-4-sqrt-2%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
$begingroup$
taking the logarithm on both sides of your equation we obatin
$$left(frac{3}{4}left(frac{ln(x)}{ln(2)}right)^2+frac{lnleft(x-frac{5}{4}right)}{ln(2)}right)ln(x)=frac{1}{2}ln(2)$$ and here you can take a numerical method
with such a method we obtain $$xapprox 2.050105808$$
answered Apr 14 '17 at 14:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
$endgroup$
$begingroup$
this is turning out to be a tricky thing. can u please solve it?
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:58
add a comment |
$begingroup$
taking the logarithm on both sides of your equation we obatin
$$left(frac{3}{4}left(frac{ln(x)}{ln(2)}right)^2+frac{lnleft(x-frac{5}{4}right)}{ln(2)}right)ln(x)=frac{1}{2}ln(2)$$ and here you can take a numerical method
with such a method we obtain $$xapprox 2.050105808$$
answered Apr 14 '17 at 14:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
$endgroup$
$begingroup$
this is turning out to be a tricky thing. can u please solve it?
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:58
add a comment |
$begingroup$
taking the logarithm on both sides of your equation we obatin
$$left(frac{3}{4}left(frac{ln(x)}{ln(2)}right)^2+frac{lnleft(x-frac{5}{4}right)}{ln(2)}right)ln(x)=frac{1}{2}ln(2)$$ and here you can take a numerical method
with such a method we obtain $$xapprox 2.050105808$$
answered Apr 14 '17 at 14:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
$endgroup$
taking the logarithm on both sides of your equation we obatin
$$left(frac{3}{4}left(frac{ln(x)}{ln(2)}right)^2+frac{lnleft(x-frac{5}{4}right)}{ln(2)}right)ln(x)=frac{1}{2}ln(2)$$ and here you can take a numerical method
with such a method we obtain $$xapprox 2.050105808$$
answered Apr 14 '17 at 14:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
edited Apr 14 '17 at 15:24
answered Apr 14 '17 at 14:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
answered Apr 14 '17 at 14:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
answered Apr 14 '17 at 14:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
75.6k42866
$begingroup$
this is turning out to be a tricky thing. can u please solve it?
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:58
add a comment |
$begingroup$
this is turning out to be a tricky thing. can u please solve it?
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:58
$begingroup$
this is turning out to be a tricky thing. can u please solve it?
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:58
$begingroup$
this is turning out to be a tricky thing. can u please solve it?
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:58
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%2f2233993%2fsolve-this-x-frac-34-log-22-x-log-2-x-5-4-sqrt-2%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
$begingroup$
Does $log^2 x$ mean $log(log x)$ or $(log x)^2$
$endgroup$
– DHMO
Apr 14 '17 at 14:39
1
$begingroup$
The title and the question have different equations in them. Which one do you mean?
$endgroup$
– AsafHaas
Apr 14 '17 at 14:40
$begingroup$
$ (log_2 x) ^ 2 $
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
@AsafHaas its the same. i hav copied it from the title.
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
By executing your idea and let $y=log_2 x$, we get $1/2=y(3y^2/4+log_2{(x-5/4)})$
$endgroup$
– didgogns
Apr 14 '17 at 14:43