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Solve Exponential equation $2^x=3$
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I want to solve this exponential equation:
$$2^x=3$$
To do this, I apply the logarithm of base 2 to both sides of the equation:
$$log_22^x = 3 implies xlog_22 = log_23 implies x = log_23$$
I would not be able to go on from here. My textbook suggests that the answer is $frac 1 9$. How? I can't really wrap my head around this even though I know this is pretty easy. Any hints?
Edit: this is what the textbook adds:
$$log_3x = -2$$
And then applies a function $3^t$ so that
$$x=3^-2= frac 1 9$$
algebra-precalculus
asked Jul 27 at 13:39
Cesare
37619
 |Â
show 4 more comments
up vote
0
down vote
favorite
I want to solve this exponential equation:
$$2^x=3$$
To do this, I apply the logarithm of base 2 to both sides of the equation:
$$log_22^x = 3 implies xlog_22 = log_23 implies x = log_23$$
I would not be able to go on from here. My textbook suggests that the answer is $frac 1 9$. How? I can't really wrap my head around this even though I know this is pretty easy. Any hints?
Edit: this is what the textbook adds:
$$log_3x = -2$$
And then applies a function $3^t$ so that
$$x=3^-2= frac 1 9$$
algebra-precalculus
asked Jul 27 at 13:39
Cesare
37619
1
You are right and text book is wrong.
– Bumblebee
Jul 27 at 13:41
1
You've gone as far as you can easily go. Analytic methods show that $log_23approx 1.584962501$. Certainly not $frac 19$ (easy to see that, in fact, it is irrational).
– lulu
Jul 27 at 13:41
Your answer is right and $1/9$ is wrong. The correct numerical value is about $1.58$. Are you sure you copied the problem correctly?
– Ethan Bolker
Jul 27 at 13:41
change book! :P
– gimusi
Jul 27 at 13:44
1
Ops, it seems that the book reports two distinct examples of logarithmic and exponential equations but it's not made clear from the text. Sorry and thanks everyone!
– Cesare
Jul 27 at 13:49
 |Â
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to solve this exponential equation:
$$2^x=3$$
To do this, I apply the logarithm of base 2 to both sides of the equation:
$$log_22^x = 3 implies xlog_22 = log_23 implies x = log_23$$
I would not be able to go on from here. My textbook suggests that the answer is $frac 1 9$. How? I can't really wrap my head around this even though I know this is pretty easy. Any hints?
Edit: this is what the textbook adds:
$$log_3x = -2$$
And then applies a function $3^t$ so that
$$x=3^-2= frac 1 9$$
algebra-precalculus
asked Jul 27 at 13:39
Cesare
37619
I want to solve this exponential equation:
$$2^x=3$$
To do this, I apply the logarithm of base 2 to both sides of the equation:
$$log_22^x = 3 implies xlog_22 = log_23 implies x = log_23$$
I would not be able to go on from here. My textbook suggests that the answer is $frac 1 9$. How? I can't really wrap my head around this even though I know this is pretty easy. Any hints?
Edit: this is what the textbook adds:
$$log_3x = -2$$
And then applies a function $3^t$ so that
$$x=3^-2= frac 1 9$$
algebra-precalculus
asked Jul 27 at 13:39
Cesare
37619
edited Jul 27 at 13:44
asked Jul 27 at 13:39
Cesare
37619
asked Jul 27 at 13:39
Cesare
37619
asked Jul 27 at 13:39
Cesare
37619
37619
1
You are right and text book is wrong.
– Bumblebee
Jul 27 at 13:41
1
You've gone as far as you can easily go. Analytic methods show that $log_23approx 1.584962501$. Certainly not $frac 19$ (easy to see that, in fact, it is irrational).
– lulu
Jul 27 at 13:41
Your answer is right and $1/9$ is wrong. The correct numerical value is about $1.58$. Are you sure you copied the problem correctly?
– Ethan Bolker
Jul 27 at 13:41
change book! :P
– gimusi
Jul 27 at 13:44
1
Ops, it seems that the book reports two distinct examples of logarithmic and exponential equations but it's not made clear from the text. Sorry and thanks everyone!
– Cesare
Jul 27 at 13:49
 |Â
show 4 more comments
1
You are right and text book is wrong.
– Bumblebee
Jul 27 at 13:41
1
You've gone as far as you can easily go. Analytic methods show that $log_23approx 1.584962501$. Certainly not $frac 19$ (easy to see that, in fact, it is irrational).
– lulu
Jul 27 at 13:41
Your answer is right and $1/9$ is wrong. The correct numerical value is about $1.58$. Are you sure you copied the problem correctly?
– Ethan Bolker
Jul 27 at 13:41
change book! :P
– gimusi
Jul 27 at 13:44
1
Ops, it seems that the book reports two distinct examples of logarithmic and exponential equations but it's not made clear from the text. Sorry and thanks everyone!
– Cesare
Jul 27 at 13:49
1
1
You are right and text book is wrong.
– Bumblebee
Jul 27 at 13:41
You are right and text book is wrong.
– Bumblebee
Jul 27 at 13:41
1
1
You've gone as far as you can easily go. Analytic methods show that $log_23approx 1.584962501$. Certainly not $frac 19$ (easy to see that, in fact, it is irrational).
– lulu
Jul 27 at 13:41
You've gone as far as you can easily go. Analytic methods show that $log_23approx 1.584962501$. Certainly not $frac 19$ (easy to see that, in fact, it is irrational).
– lulu
Jul 27 at 13:41
Your answer is right and $1/9$ is wrong. The correct numerical value is about $1.58$. Are you sure you copied the problem correctly?
– Ethan Bolker
Jul 27 at 13:41
Your answer is right and $1/9$ is wrong. The correct numerical value is about $1.58$. Are you sure you copied the problem correctly?
– Ethan Bolker
Jul 27 at 13:41
change book! :P
– gimusi
Jul 27 at 13:44
change book! :P
– gimusi
Jul 27 at 13:44
1
1
Ops, it seems that the book reports two distinct examples of logarithmic and exponential equations but it's not made clear from the text. Sorry and thanks everyone!
– Cesare
Jul 27 at 13:49
Ops, it seems that the book reports two distinct examples of logarithmic and exponential equations but it's not made clear from the text. Sorry and thanks everyone!
– Cesare
Jul 27 at 13:49
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Yes your solution is absolutely correct, indeed by definition
$$2^x=2^(log_2 3)=3$$
That seems the solution to
$$log_3 x=-2 implies 3^log_3 x=3^-2 implies x=frac19 $$
answered Jul 27 at 13:42
gimusi
64.9k73583
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yes your solution is absolutely correct, indeed by definition
$$2^x=2^(log_2 3)=3$$
That seems the solution to
$$log_3 x=-2 implies 3^log_3 x=3^-2 implies x=frac19 $$
answered Jul 27 at 13:42
gimusi
64.9k73583
add a comment |Â
up vote
2
down vote
accepted
Yes your solution is absolutely correct, indeed by definition
$$2^x=2^(log_2 3)=3$$
That seems the solution to
$$log_3 x=-2 implies 3^log_3 x=3^-2 implies x=frac19 $$
answered Jul 27 at 13:42
gimusi
64.9k73583
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yes your solution is absolutely correct, indeed by definition
$$2^x=2^(log_2 3)=3$$
That seems the solution to
$$log_3 x=-2 implies 3^log_3 x=3^-2 implies x=frac19 $$
answered Jul 27 at 13:42
gimusi
64.9k73583
Yes your solution is absolutely correct, indeed by definition
$$2^x=2^(log_2 3)=3$$
That seems the solution to
$$log_3 x=-2 implies 3^log_3 x=3^-2 implies x=frac19 $$
answered Jul 27 at 13:42
gimusi
64.9k73583
answered Jul 27 at 13:42
gimusi
64.9k73583
answered Jul 27 at 13:42
gimusi
64.9k73583
answered Jul 27 at 13:42
gimusi
64.9k73583
64.9k73583
add a comment |Â
add a comment |Â
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1
You are right and text book is wrong.
– Bumblebee
Jul 27 at 13:41
1
You've gone as far as you can easily go. Analytic methods show that $log_23approx 1.584962501$. Certainly not $frac 19$ (easy to see that, in fact, it is irrational).
– lulu
Jul 27 at 13:41
Your answer is right and $1/9$ is wrong. The correct numerical value is about $1.58$. Are you sure you copied the problem correctly?
– Ethan Bolker
Jul 27 at 13:41
change book! :P
– gimusi
Jul 27 at 13:44
1
Ops, it seems that the book reports two distinct examples of logarithmic and exponential equations but it's not made clear from the text. Sorry and thanks everyone!
– Cesare
Jul 27 at 13:49