Calculate the factor increase for y=ln x
$begingroup$
Question:
Suppose $x >1$ and $x$ increases by a factor of $5$. By what factor will $y$ increase given $y = ln(x)$?
My answer:
So the increase would be $$frac{(ln(5x) - ln x)}{ln x} times 100%$$
And then $$frac{( ln 5 + ln x -ln x )}{ln x}times 100 %$$
But thats clearly wrong because $ln x$ get cancelled in the numerator but the denominator is $ln x$
logarithms
edited Dec 13 '18 at 19:33
Emilio Novati
51.6k43473
asked Dec 13 '18 at 17:22
LJacobLJacob
61
$endgroup$
add a comment |
$begingroup$
Question:
Suppose $x >1$ and $x$ increases by a factor of $5$. By what factor will $y$ increase given $y = ln(x)$?
My answer:
So the increase would be $$frac{(ln(5x) - ln x)}{ln x} times 100%$$
And then $$frac{( ln 5 + ln x -ln x )}{ln x}times 100 %$$
But thats clearly wrong because $ln x$ get cancelled in the numerator but the denominator is $ln x$
logarithms
edited Dec 13 '18 at 19:33
Emilio Novati
51.6k43473
asked Dec 13 '18 at 17:22
LJacobLJacob
61
$endgroup$
$begingroup$
Why must that be wrong? The factor of increase in $y$ will not be independent of $x$. Try some numerical examples. (Please edit the question to reformat with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 13 '18 at 17:26
add a comment |
$begingroup$
Question:
Suppose $x >1$ and $x$ increases by a factor of $5$. By what factor will $y$ increase given $y = ln(x)$?
My answer:
So the increase would be $$frac{(ln(5x) - ln x)}{ln x} times 100%$$
And then $$frac{( ln 5 + ln x -ln x )}{ln x}times 100 %$$
But thats clearly wrong because $ln x$ get cancelled in the numerator but the denominator is $ln x$
logarithms
edited Dec 13 '18 at 19:33
Emilio Novati
51.6k43473
asked Dec 13 '18 at 17:22
LJacobLJacob
61
$endgroup$
Question:
Suppose $x >1$ and $x$ increases by a factor of $5$. By what factor will $y$ increase given $y = ln(x)$?
My answer:
So the increase would be $$frac{(ln(5x) - ln x)}{ln x} times 100%$$
And then $$frac{( ln 5 + ln x -ln x )}{ln x}times 100 %$$
But thats clearly wrong because $ln x$ get cancelled in the numerator but the denominator is $ln x$
logarithms
logarithms
edited Dec 13 '18 at 19:33
Emilio Novati
51.6k43473
asked Dec 13 '18 at 17:22
LJacobLJacob
61
edited Dec 13 '18 at 19:33
Emilio Novati
51.6k43473
asked Dec 13 '18 at 17:22
LJacobLJacob
61
edited Dec 13 '18 at 19:33
Emilio Novati
51.6k43473
edited Dec 13 '18 at 19:33
Emilio Novati
51.6k43473
edited Dec 13 '18 at 19:33
Emilio Novati
51.6k43473
51.6k43473
asked Dec 13 '18 at 17:22
LJacobLJacob
61
asked Dec 13 '18 at 17:22
LJacobLJacob
61
asked Dec 13 '18 at 17:22
LJacobLJacob
61
61
$begingroup$
Why must that be wrong? The factor of increase in $y$ will not be independent of $x$. Try some numerical examples. (Please edit the question to reformat with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 13 '18 at 17:26
add a comment |
$begingroup$
Why must that be wrong? The factor of increase in $y$ will not be independent of $x$. Try some numerical examples. (Please edit the question to reformat with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 13 '18 at 17:26
$begingroup$
Why must that be wrong? The factor of increase in $y$ will not be independent of $x$. Try some numerical examples. (Please edit the question to reformat with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 13 '18 at 17:26
$begingroup$
Why must that be wrong? The factor of increase in $y$ will not be independent of $x$. Try some numerical examples. (Please edit the question to reformat with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 13 '18 at 17:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The increase in $y$, is just given by $ Delta y=ln(5x)-ln x=ln 5$, which is independent of the value of $x$.
The relative increase in $y$ is given by $frac{Delta y}y=ln5/ln x$, which is not independent of $x$, and can also be expressed as a percentage.
answered Dec 13 '18 at 17:31
Shubham JohriShubham Johri
4,666717
$endgroup$
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%2f3038331%2fcalculate-the-factor-increase-for-y-ln-x%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
$begingroup$
The increase in $y$, is just given by $ Delta y=ln(5x)-ln x=ln 5$, which is independent of the value of $x$.
The relative increase in $y$ is given by $frac{Delta y}y=ln5/ln x$, which is not independent of $x$, and can also be expressed as a percentage.
answered Dec 13 '18 at 17:31
Shubham JohriShubham Johri
4,666717
$endgroup$
add a comment |
$begingroup$
The increase in $y$, is just given by $ Delta y=ln(5x)-ln x=ln 5$, which is independent of the value of $x$.
The relative increase in $y$ is given by $frac{Delta y}y=ln5/ln x$, which is not independent of $x$, and can also be expressed as a percentage.
answered Dec 13 '18 at 17:31
Shubham JohriShubham Johri
4,666717
$endgroup$
add a comment |
$begingroup$
The increase in $y$, is just given by $ Delta y=ln(5x)-ln x=ln 5$, which is independent of the value of $x$.
The relative increase in $y$ is given by $frac{Delta y}y=ln5/ln x$, which is not independent of $x$, and can also be expressed as a percentage.
answered Dec 13 '18 at 17:31
Shubham JohriShubham Johri
4,666717
$endgroup$
The increase in $y$, is just given by $ Delta y=ln(5x)-ln x=ln 5$, which is independent of the value of $x$.
The relative increase in $y$ is given by $frac{Delta y}y=ln5/ln x$, which is not independent of $x$, and can also be expressed as a percentage.
answered Dec 13 '18 at 17:31
Shubham JohriShubham Johri
4,666717
answered Dec 13 '18 at 17:31
Shubham JohriShubham Johri
4,666717
answered Dec 13 '18 at 17:31
Shubham JohriShubham Johri
4,666717
answered Dec 13 '18 at 17:31
Shubham JohriShubham Johri
4,666717
4,666717
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%2f3038331%2fcalculate-the-factor-increase-for-y-ln-x%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
$begingroup$
Why must that be wrong? The factor of increase in $y$ will not be independent of $x$. Try some numerical examples. (Please edit the question to reformat with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 13 '18 at 17:26