Middle School Log question
$$left(frac 43right)^x=frac{8sqrt3}{9}$$
I've been trying to solve this question. I know that the answer is $3/2$ by guess and check, but I want to know how to do it by actually solving it.
logarithms
add a comment |
$$left(frac 43right)^x=frac{8sqrt3}{9}$$
I've been trying to solve this question. I know that the answer is $3/2$ by guess and check, but I want to know how to do it by actually solving it.
logarithms
1
If you have an acceptable answer, accept it by clicking the check mark on that answer.
– Sean Roberson
27 mins ago
add a comment |
$$left(frac 43right)^x=frac{8sqrt3}{9}$$
I've been trying to solve this question. I know that the answer is $3/2$ by guess and check, but I want to know how to do it by actually solving it.
logarithms
$$left(frac 43right)^x=frac{8sqrt3}{9}$$
I've been trying to solve this question. I know that the answer is $3/2$ by guess and check, but I want to know how to do it by actually solving it.
logarithms
logarithms
edited 14 mins ago
Rócherz
2,7362721
2,7362721
asked 47 mins ago
Toylatte
133
133
1
If you have an acceptable answer, accept it by clicking the check mark on that answer.
– Sean Roberson
27 mins ago
add a comment |
1
If you have an acceptable answer, accept it by clicking the check mark on that answer.
– Sean Roberson
27 mins ago
1
1
If you have an acceptable answer, accept it by clicking the check mark on that answer.
– Sean Roberson
27 mins ago
If you have an acceptable answer, accept it by clicking the check mark on that answer.
– Sean Roberson
27 mins ago
add a comment |
2 Answers
2
active
oldest
votes
I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
$$
left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
$$
Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!
Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.
New contributor
Thanks it helped
– Toylatte
39 mins ago
@Toylatte You are welcome!
– ImNotTheGuy
39 mins ago
add a comment |
$$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$
Note that the function $f(x)=left( frac43right)^x$ is an increasing function.
Hence $x=1.5$.
Alternatively, taking logarithm on both sides.
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%2f3052593%2fmiddle-school-log-question%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
$$
left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
$$
Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!
Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.
New contributor
Thanks it helped
– Toylatte
39 mins ago
@Toylatte You are welcome!
– ImNotTheGuy
39 mins ago
add a comment |
I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
$$
left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
$$
Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!
Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.
New contributor
Thanks it helped
– Toylatte
39 mins ago
@Toylatte You are welcome!
– ImNotTheGuy
39 mins ago
add a comment |
I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
$$
left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
$$
Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!
Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.
New contributor
I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
$$
left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
$$
Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!
Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.
New contributor
New contributor
answered 42 mins ago
ImNotTheGuy
3364
3364
New contributor
New contributor
Thanks it helped
– Toylatte
39 mins ago
@Toylatte You are welcome!
– ImNotTheGuy
39 mins ago
add a comment |
Thanks it helped
– Toylatte
39 mins ago
@Toylatte You are welcome!
– ImNotTheGuy
39 mins ago
Thanks it helped
– Toylatte
39 mins ago
Thanks it helped
– Toylatte
39 mins ago
@Toylatte You are welcome!
– ImNotTheGuy
39 mins ago
@Toylatte You are welcome!
– ImNotTheGuy
39 mins ago
add a comment |
$$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$
Note that the function $f(x)=left( frac43right)^x$ is an increasing function.
Hence $x=1.5$.
Alternatively, taking logarithm on both sides.
add a comment |
$$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$
Note that the function $f(x)=left( frac43right)^x$ is an increasing function.
Hence $x=1.5$.
Alternatively, taking logarithm on both sides.
add a comment |
$$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$
Note that the function $f(x)=left( frac43right)^x$ is an increasing function.
Hence $x=1.5$.
Alternatively, taking logarithm on both sides.
$$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$
Note that the function $f(x)=left( frac43right)^x$ is an increasing function.
Hence $x=1.5$.
Alternatively, taking logarithm on both sides.
answered 42 mins ago
Siong Thye Goh
98.8k1464116
98.8k1464116
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3052593%2fmiddle-school-log-question%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
1
If you have an acceptable answer, accept it by clicking the check mark on that answer.
– Sean Roberson
27 mins ago