How do I solve the equation: $log_2(x-1)+log_2(3)=log_{16}(2x)$ [closed]
How do I solve the equation: $log_2(x-1)+log_2(3)=log_{16}(2x)$
I know the basics of logarithm. I solved up to the point: $(x-1)^4 cdot 81=2x$. I don't know if I did the right steps to get to this point.
calculus
edited Dec 11 '18 at 3:45
Zach Langley
9691019
asked Dec 11 '18 at 3:36
User231
135
closed as off-topic by Nosrati, user10354138, Shaun, ncmathsadist, José Carlos Santos Dec 11 '18 at 15:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, user10354138, Shaun, ncmathsadist, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
How do I solve the equation: $log_2(x-1)+log_2(3)=log_{16}(2x)$
I know the basics of logarithm. I solved up to the point: $(x-1)^4 cdot 81=2x$. I don't know if I did the right steps to get to this point.
calculus
edited Dec 11 '18 at 3:45
Zach Langley
9691019
asked Dec 11 '18 at 3:36
User231
135
closed as off-topic by Nosrati, user10354138, Shaun, ncmathsadist, José Carlos Santos Dec 11 '18 at 15:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, user10354138, Shaun, ncmathsadist, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Why don't you show us the steps you used to get to that point so we can see whether you have done it correctly or not?
– Mattos
Dec 11 '18 at 3:39
I'm not used to the keyboard because I'm new to MathStack Exchange and I can't take a picture. It would take a long time to write out all the steps.
– User231
Dec 11 '18 at 3:45
6
I'm not sure what your point is. Then it takes a long time.
– Mattos
Dec 11 '18 at 3:46
Try applying the change of base formula. And, if you are going to routinely use this forum, you should learn its lingua franca. There is an excellent tutorial on the site.
– ncmathsadist
Dec 11 '18 at 12:58
add a comment |
How do I solve the equation: $log_2(x-1)+log_2(3)=log_{16}(2x)$
I know the basics of logarithm. I solved up to the point: $(x-1)^4 cdot 81=2x$. I don't know if I did the right steps to get to this point.
calculus
edited Dec 11 '18 at 3:45
Zach Langley
9691019
asked Dec 11 '18 at 3:36
User231
135
How do I solve the equation: $log_2(x-1)+log_2(3)=log_{16}(2x)$
I know the basics of logarithm. I solved up to the point: $(x-1)^4 cdot 81=2x$. I don't know if I did the right steps to get to this point.
calculus
calculus
edited Dec 11 '18 at 3:45
Zach Langley
9691019
asked Dec 11 '18 at 3:36
User231
135
edited Dec 11 '18 at 3:45
Zach Langley
9691019
asked Dec 11 '18 at 3:36
User231
135
edited Dec 11 '18 at 3:45
Zach Langley
9691019
edited Dec 11 '18 at 3:45
Zach Langley
9691019
edited Dec 11 '18 at 3:45
Zach Langley
9691019
9691019
asked Dec 11 '18 at 3:36
User231
135
asked Dec 11 '18 at 3:36
User231
135
asked Dec 11 '18 at 3:36
User231
135
135
closed as off-topic by Nosrati, user10354138, Shaun, ncmathsadist, José Carlos Santos Dec 11 '18 at 15:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, user10354138, Shaun, ncmathsadist, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Nosrati, user10354138, Shaun, ncmathsadist, José Carlos Santos Dec 11 '18 at 15:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, user10354138, Shaun, ncmathsadist, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Why don't you show us the steps you used to get to that point so we can see whether you have done it correctly or not?
– Mattos
Dec 11 '18 at 3:39
I'm not used to the keyboard because I'm new to MathStack Exchange and I can't take a picture. It would take a long time to write out all the steps.
– User231
Dec 11 '18 at 3:45
6
I'm not sure what your point is. Then it takes a long time.
– Mattos
Dec 11 '18 at 3:46
Try applying the change of base formula. And, if you are going to routinely use this forum, you should learn its lingua franca. There is an excellent tutorial on the site.
– ncmathsadist
Dec 11 '18 at 12:58
add a comment |
1
Why don't you show us the steps you used to get to that point so we can see whether you have done it correctly or not?
– Mattos
Dec 11 '18 at 3:39
I'm not used to the keyboard because I'm new to MathStack Exchange and I can't take a picture. It would take a long time to write out all the steps.
– User231
Dec 11 '18 at 3:45
6
I'm not sure what your point is. Then it takes a long time.
– Mattos
Dec 11 '18 at 3:46
Try applying the change of base formula. And, if you are going to routinely use this forum, you should learn its lingua franca. There is an excellent tutorial on the site.
– ncmathsadist
Dec 11 '18 at 12:58
1
1
Why don't you show us the steps you used to get to that point so we can see whether you have done it correctly or not?
– Mattos
Dec 11 '18 at 3:39
Why don't you show us the steps you used to get to that point so we can see whether you have done it correctly or not?
– Mattos
Dec 11 '18 at 3:39
I'm not used to the keyboard because I'm new to MathStack Exchange and I can't take a picture. It would take a long time to write out all the steps.
– User231
Dec 11 '18 at 3:45
I'm not used to the keyboard because I'm new to MathStack Exchange and I can't take a picture. It would take a long time to write out all the steps.
– User231
Dec 11 '18 at 3:45
6
6
I'm not sure what your point is. Then it takes a long time.
– Mattos
Dec 11 '18 at 3:46
I'm not sure what your point is. Then it takes a long time.
– Mattos
Dec 11 '18 at 3:46
Try applying the change of base formula. And, if you are going to routinely use this forum, you should learn its lingua franca. There is an excellent tutorial on the site.
– ncmathsadist
Dec 11 '18 at 12:58
Try applying the change of base formula. And, if you are going to routinely use this forum, you should learn its lingua franca. There is an excellent tutorial on the site.
– ncmathsadist
Dec 11 '18 at 12:58
add a comment |
1 Answer
1
active
oldest
votes
By the log-of-product formula we have,
$$log_2(x-1)+log_2(3)=log_2(3x-3)$$
and by change of basis we have,
$$log_{16}(2x)=frac{log_{2}(2x)}{log_{2}16}=frac{1}{4}log_2 (2x)$$
so we have the equality,
$$log_2(3x-3)=frac{1}{4}log_2(2x)$$
raise to power 2 we get,
$$3x-3=(2^{log_2(2x)})^{frac{1}{4}}=(2x)^{frac{1}{4}}$$
can you solve from here? make sure to plug in whatever solutions you get back into the original equation; you might get extraneous (false) "solutions". Note that we require $x>1$ in order for $log_2(3x-3)$ to exist (I'm assuming you're not concerned with complex numbers here?
answered Dec 11 '18 at 3:46
clmundergrad
1665
Thank you! I'm somewhat concerned with complex numbers though.
– User231
Dec 11 '18 at 4:05
@User231 If you're concerned about complex numbers, you have to specify which branch of the complex logarithm function is being used. It's usually not worth the trouble.
– eyeballfrog
Dec 11 '18 at 5:17
I haven't done complex analysis in a while, I just remember logarithms being a pain since, as mentioned above, you have to specify what branch is being used. This has to do with the fact that exp(x)=exp(x+2pi*i)
– clmundergrad
Dec 11 '18 at 7:47
Thank you! So i set equation to zero to solve it? Iwas judt unsure of that part.
– User231
Dec 11 '18 at 11:44
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
By the log-of-product formula we have,
$$log_2(x-1)+log_2(3)=log_2(3x-3)$$
and by change of basis we have,
$$log_{16}(2x)=frac{log_{2}(2x)}{log_{2}16}=frac{1}{4}log_2 (2x)$$
so we have the equality,
$$log_2(3x-3)=frac{1}{4}log_2(2x)$$
raise to power 2 we get,
$$3x-3=(2^{log_2(2x)})^{frac{1}{4}}=(2x)^{frac{1}{4}}$$
can you solve from here? make sure to plug in whatever solutions you get back into the original equation; you might get extraneous (false) "solutions". Note that we require $x>1$ in order for $log_2(3x-3)$ to exist (I'm assuming you're not concerned with complex numbers here?
answered Dec 11 '18 at 3:46
clmundergrad
1665
Thank you! I'm somewhat concerned with complex numbers though.
– User231
Dec 11 '18 at 4:05
@User231 If you're concerned about complex numbers, you have to specify which branch of the complex logarithm function is being used. It's usually not worth the trouble.
– eyeballfrog
Dec 11 '18 at 5:17
I haven't done complex analysis in a while, I just remember logarithms being a pain since, as mentioned above, you have to specify what branch is being used. This has to do with the fact that exp(x)=exp(x+2pi*i)
– clmundergrad
Dec 11 '18 at 7:47
Thank you! So i set equation to zero to solve it? Iwas judt unsure of that part.
– User231
Dec 11 '18 at 11:44
add a comment |
By the log-of-product formula we have,
$$log_2(x-1)+log_2(3)=log_2(3x-3)$$
and by change of basis we have,
$$log_{16}(2x)=frac{log_{2}(2x)}{log_{2}16}=frac{1}{4}log_2 (2x)$$
so we have the equality,
$$log_2(3x-3)=frac{1}{4}log_2(2x)$$
raise to power 2 we get,
$$3x-3=(2^{log_2(2x)})^{frac{1}{4}}=(2x)^{frac{1}{4}}$$
can you solve from here? make sure to plug in whatever solutions you get back into the original equation; you might get extraneous (false) "solutions". Note that we require $x>1$ in order for $log_2(3x-3)$ to exist (I'm assuming you're not concerned with complex numbers here?
answered Dec 11 '18 at 3:46
clmundergrad
1665
Thank you! I'm somewhat concerned with complex numbers though.
– User231
Dec 11 '18 at 4:05
@User231 If you're concerned about complex numbers, you have to specify which branch of the complex logarithm function is being used. It's usually not worth the trouble.
– eyeballfrog
Dec 11 '18 at 5:17
I haven't done complex analysis in a while, I just remember logarithms being a pain since, as mentioned above, you have to specify what branch is being used. This has to do with the fact that exp(x)=exp(x+2pi*i)
– clmundergrad
Dec 11 '18 at 7:47
Thank you! So i set equation to zero to solve it? Iwas judt unsure of that part.
– User231
Dec 11 '18 at 11:44
add a comment |
By the log-of-product formula we have,
$$log_2(x-1)+log_2(3)=log_2(3x-3)$$
and by change of basis we have,
$$log_{16}(2x)=frac{log_{2}(2x)}{log_{2}16}=frac{1}{4}log_2 (2x)$$
so we have the equality,
$$log_2(3x-3)=frac{1}{4}log_2(2x)$$
raise to power 2 we get,
$$3x-3=(2^{log_2(2x)})^{frac{1}{4}}=(2x)^{frac{1}{4}}$$
can you solve from here? make sure to plug in whatever solutions you get back into the original equation; you might get extraneous (false) "solutions". Note that we require $x>1$ in order for $log_2(3x-3)$ to exist (I'm assuming you're not concerned with complex numbers here?
answered Dec 11 '18 at 3:46
clmundergrad
1665
By the log-of-product formula we have,
$$log_2(x-1)+log_2(3)=log_2(3x-3)$$
and by change of basis we have,
$$log_{16}(2x)=frac{log_{2}(2x)}{log_{2}16}=frac{1}{4}log_2 (2x)$$
so we have the equality,
$$log_2(3x-3)=frac{1}{4}log_2(2x)$$
raise to power 2 we get,
$$3x-3=(2^{log_2(2x)})^{frac{1}{4}}=(2x)^{frac{1}{4}}$$
can you solve from here? make sure to plug in whatever solutions you get back into the original equation; you might get extraneous (false) "solutions". Note that we require $x>1$ in order for $log_2(3x-3)$ to exist (I'm assuming you're not concerned with complex numbers here?
answered Dec 11 '18 at 3:46
clmundergrad
1665
answered Dec 11 '18 at 3:46
clmundergrad
1665
answered Dec 11 '18 at 3:46
clmundergrad
1665
answered Dec 11 '18 at 3:46
clmundergrad
1665
1665
Thank you! I'm somewhat concerned with complex numbers though.
– User231
Dec 11 '18 at 4:05
@User231 If you're concerned about complex numbers, you have to specify which branch of the complex logarithm function is being used. It's usually not worth the trouble.
– eyeballfrog
Dec 11 '18 at 5:17
I haven't done complex analysis in a while, I just remember logarithms being a pain since, as mentioned above, you have to specify what branch is being used. This has to do with the fact that exp(x)=exp(x+2pi*i)
– clmundergrad
Dec 11 '18 at 7:47
Thank you! So i set equation to zero to solve it? Iwas judt unsure of that part.
– User231
Dec 11 '18 at 11:44
add a comment |
Thank you! I'm somewhat concerned with complex numbers though.
– User231
Dec 11 '18 at 4:05
@User231 If you're concerned about complex numbers, you have to specify which branch of the complex logarithm function is being used. It's usually not worth the trouble.
– eyeballfrog
Dec 11 '18 at 5:17
I haven't done complex analysis in a while, I just remember logarithms being a pain since, as mentioned above, you have to specify what branch is being used. This has to do with the fact that exp(x)=exp(x+2pi*i)
– clmundergrad
Dec 11 '18 at 7:47
Thank you! So i set equation to zero to solve it? Iwas judt unsure of that part.
– User231
Dec 11 '18 at 11:44
Thank you! I'm somewhat concerned with complex numbers though.
– User231
Dec 11 '18 at 4:05
Thank you! I'm somewhat concerned with complex numbers though.
– User231
Dec 11 '18 at 4:05
@User231 If you're concerned about complex numbers, you have to specify which branch of the complex logarithm function is being used. It's usually not worth the trouble.
– eyeballfrog
Dec 11 '18 at 5:17
@User231 If you're concerned about complex numbers, you have to specify which branch of the complex logarithm function is being used. It's usually not worth the trouble.
– eyeballfrog
Dec 11 '18 at 5:17
I haven't done complex analysis in a while, I just remember logarithms being a pain since, as mentioned above, you have to specify what branch is being used. This has to do with the fact that exp(x)=exp(x+2pi*i)
– clmundergrad
Dec 11 '18 at 7:47
I haven't done complex analysis in a while, I just remember logarithms being a pain since, as mentioned above, you have to specify what branch is being used. This has to do with the fact that exp(x)=exp(x+2pi*i)
– clmundergrad
Dec 11 '18 at 7:47
Thank you! So i set equation to zero to solve it? Iwas judt unsure of that part.
– User231
Dec 11 '18 at 11:44
Thank you! So i set equation to zero to solve it? Iwas judt unsure of that part.
– User231
Dec 11 '18 at 11:44
add a comment |
1
Why don't you show us the steps you used to get to that point so we can see whether you have done it correctly or not?
– Mattos
Dec 11 '18 at 3:39
I'm not used to the keyboard because I'm new to MathStack Exchange and I can't take a picture. It would take a long time to write out all the steps.
– User231
Dec 11 '18 at 3:45
6
I'm not sure what your point is. Then it takes a long time.
– Mattos
Dec 11 '18 at 3:46
Try applying the change of base formula. And, if you are going to routinely use this forum, you should learn its lingua franca. There is an excellent tutorial on the site.
– ncmathsadist
Dec 11 '18 at 12:58