How do you solve the equation $0.5^x = 2^x + 3$?
$begingroup$
I need help with the following problem: $0.5^x = 2^x + 3$
I know the answer is -1.72, but I have to explain step by step how to solve it and I'm not sure how. I know you're supposed to take the log of each side, but I don't know what to do with the 3. How do you solve this problem?
logarithms
edited Jan 11 '16 at 13:59
Adrian
5,2291135
asked Jan 11 '16 at 13:56
JohnJohn
1
$endgroup$
add a comment |
$begingroup$
I need help with the following problem: $0.5^x = 2^x + 3$
I know the answer is -1.72, but I have to explain step by step how to solve it and I'm not sure how. I know you're supposed to take the log of each side, but I don't know what to do with the 3. How do you solve this problem?
logarithms
edited Jan 11 '16 at 13:59
Adrian
5,2291135
asked Jan 11 '16 at 13:56
JohnJohn
1
$endgroup$
add a comment |
$begingroup$
I need help with the following problem: $0.5^x = 2^x + 3$
I know the answer is -1.72, but I have to explain step by step how to solve it and I'm not sure how. I know you're supposed to take the log of each side, but I don't know what to do with the 3. How do you solve this problem?
logarithms
edited Jan 11 '16 at 13:59
Adrian
5,2291135
asked Jan 11 '16 at 13:56
JohnJohn
1
$endgroup$
I need help with the following problem: $0.5^x = 2^x + 3$
I know the answer is -1.72, but I have to explain step by step how to solve it and I'm not sure how. I know you're supposed to take the log of each side, but I don't know what to do with the 3. How do you solve this problem?
logarithms
logarithms
edited Jan 11 '16 at 13:59
Adrian
5,2291135
asked Jan 11 '16 at 13:56
JohnJohn
1
edited Jan 11 '16 at 13:59
Adrian
5,2291135
asked Jan 11 '16 at 13:56
JohnJohn
1
edited Jan 11 '16 at 13:59
Adrian
5,2291135
edited Jan 11 '16 at 13:59
Adrian
5,2291135
edited Jan 11 '16 at 13:59
Adrian
5,2291135
5,2291135
asked Jan 11 '16 at 13:56
JohnJohn
1
asked Jan 11 '16 at 13:56
JohnJohn
1
asked Jan 11 '16 at 13:56
JohnJohn
1
1
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Rewrite the left-hand side to $frac{1}{2^x}$, and then set $y=2^x$. This gives you a quadratic equation in $y$, which you can solve; finally $x=log_2 y$.
answered Jan 11 '16 at 13:57
Henning MakholmHenning Makholm
239k17304541
$endgroup$
add a comment |
$begingroup$
The equation is $1/2^x=2^x+3$ which gives $2^{2x}+3.2^x-1=0$
Let $a=2^x$ then $a^2+3a-1=0$. Solve for $a$, take the positive root, and solve for $x$ by setting $x=log_2(a)$.
answered Jan 11 '16 at 13:58
Landon CarterLandon Carter
7,34711543
$endgroup$
add a comment |
$begingroup$
you should be knowing $0.5^x=left(frac 12right)^{x}=frac 1{2^x}$, hence
$$0.5^{x}=2^x+3$$
$$frac 1{2^x}=2^x+3$$
$$(2^{x})^2+3cdot 2^x-1=0$$
Now, solving the quadratic equation in terms of $2^x$ as follows
$$2^x=frac{-3pmsqrt{(3)^2-4(1)(-1)}}{2(1)}=frac{-3pmsqrt{13}}{2}$$
But $2^x>0 forall xin mathbb{R}$ hence $$2^x=frac{-3+sqrt{13}}{2}$$
$$ln 2^x=lnleft(frac{-3+sqrt{13}}{2}right)$$
$$xln 2=lnleft(frac{sqrt{13}-3}{2}right)$$
$$x=color{red}{frac{lnleft(frac{sqrt{13}-3}{2}right)}{ln(2)}}$$
answered Jan 11 '16 at 14:46
Harish Chandra RajpootHarish Chandra Rajpoot
29.5k103671
$endgroup$
add a comment |
$begingroup$
$0.5^x=2^x +3$
$(frac{1}{2})^{x}=2^x + 3$
$frac{1}{2^{x}}=2^x + 3 $ $/2^x$
$1= 2^2x + 3* 2^x$
Now you make substitution, in these exercises method that is often used.
$y=2^x$
Our eqation looks like this:
$1=y^2 +3y$
$y^2 +3y-1=0$
$y=frac{-3pm sqrt{9+4}}{2}$
$y=frac{-3pm sqrt{13}}{2}$
$y_{1}=frac{-3-sqrt{13}}{2}$ or $ y_{2}=frac{-3+ sqrt{13}}{2}$
As we defined y= 2x, we know that it can not be negative number,so y1 is incorrect, and we continue only with y2.
$y=2^x$
$2^x=frac{-3+ sqrt{13}}{2} /log_{2}$
(log2(2x)=x)
$x=log_{2}frac{-3+sqrt{13}}{2}$
put it in calculator and you get :
x= -1.723
answered Jan 11 '16 at 14:43
florenceflorence
264
$endgroup$
$begingroup$
Welcome to Math Stackexchange! Did you know that you can type your math by surrounding it by$symbols?
$endgroup$
– JnxF
Jan 11 '16 at 14:56
add a comment |
$begingroup$
This may be a approach to solve it, hope this helps.
Thank you!
answered Jan 7 at 6:47
user629353user629353
1187
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Rewrite the left-hand side to $frac{1}{2^x}$, and then set $y=2^x$. This gives you a quadratic equation in $y$, which you can solve; finally $x=log_2 y$.
answered Jan 11 '16 at 13:57
Henning MakholmHenning Makholm
239k17304541
$endgroup$
add a comment |
$begingroup$
Rewrite the left-hand side to $frac{1}{2^x}$, and then set $y=2^x$. This gives you a quadratic equation in $y$, which you can solve; finally $x=log_2 y$.
answered Jan 11 '16 at 13:57
Henning MakholmHenning Makholm
239k17304541
$endgroup$
add a comment |
$begingroup$
Rewrite the left-hand side to $frac{1}{2^x}$, and then set $y=2^x$. This gives you a quadratic equation in $y$, which you can solve; finally $x=log_2 y$.
answered Jan 11 '16 at 13:57
Henning MakholmHenning Makholm
239k17304541
$endgroup$
Rewrite the left-hand side to $frac{1}{2^x}$, and then set $y=2^x$. This gives you a quadratic equation in $y$, which you can solve; finally $x=log_2 y$.
answered Jan 11 '16 at 13:57
Henning MakholmHenning Makholm
239k17304541
answered Jan 11 '16 at 13:57
Henning MakholmHenning Makholm
239k17304541
answered Jan 11 '16 at 13:57
Henning MakholmHenning Makholm
239k17304541
answered Jan 11 '16 at 13:57
Henning MakholmHenning Makholm
239k17304541
239k17304541
add a comment |
add a comment |
$begingroup$
The equation is $1/2^x=2^x+3$ which gives $2^{2x}+3.2^x-1=0$
Let $a=2^x$ then $a^2+3a-1=0$. Solve for $a$, take the positive root, and solve for $x$ by setting $x=log_2(a)$.
answered Jan 11 '16 at 13:58
Landon CarterLandon Carter
7,34711543
$endgroup$
add a comment |
$begingroup$
The equation is $1/2^x=2^x+3$ which gives $2^{2x}+3.2^x-1=0$
Let $a=2^x$ then $a^2+3a-1=0$. Solve for $a$, take the positive root, and solve for $x$ by setting $x=log_2(a)$.
answered Jan 11 '16 at 13:58
Landon CarterLandon Carter
7,34711543
$endgroup$
add a comment |
$begingroup$
The equation is $1/2^x=2^x+3$ which gives $2^{2x}+3.2^x-1=0$
Let $a=2^x$ then $a^2+3a-1=0$. Solve for $a$, take the positive root, and solve for $x$ by setting $x=log_2(a)$.
answered Jan 11 '16 at 13:58
Landon CarterLandon Carter
7,34711543
$endgroup$
The equation is $1/2^x=2^x+3$ which gives $2^{2x}+3.2^x-1=0$
Let $a=2^x$ then $a^2+3a-1=0$. Solve for $a$, take the positive root, and solve for $x$ by setting $x=log_2(a)$.
answered Jan 11 '16 at 13:58
Landon CarterLandon Carter
7,34711543
answered Jan 11 '16 at 13:58
Landon CarterLandon Carter
7,34711543
answered Jan 11 '16 at 13:58
Landon CarterLandon Carter
7,34711543
answered Jan 11 '16 at 13:58
Landon CarterLandon Carter
7,34711543
7,34711543
add a comment |
add a comment |
$begingroup$
you should be knowing $0.5^x=left(frac 12right)^{x}=frac 1{2^x}$, hence
$$0.5^{x}=2^x+3$$
$$frac 1{2^x}=2^x+3$$
$$(2^{x})^2+3cdot 2^x-1=0$$
Now, solving the quadratic equation in terms of $2^x$ as follows
$$2^x=frac{-3pmsqrt{(3)^2-4(1)(-1)}}{2(1)}=frac{-3pmsqrt{13}}{2}$$
But $2^x>0 forall xin mathbb{R}$ hence $$2^x=frac{-3+sqrt{13}}{2}$$
$$ln 2^x=lnleft(frac{-3+sqrt{13}}{2}right)$$
$$xln 2=lnleft(frac{sqrt{13}-3}{2}right)$$
$$x=color{red}{frac{lnleft(frac{sqrt{13}-3}{2}right)}{ln(2)}}$$
answered Jan 11 '16 at 14:46
Harish Chandra RajpootHarish Chandra Rajpoot
29.5k103671
$endgroup$
add a comment |
$begingroup$
you should be knowing $0.5^x=left(frac 12right)^{x}=frac 1{2^x}$, hence
$$0.5^{x}=2^x+3$$
$$frac 1{2^x}=2^x+3$$
$$(2^{x})^2+3cdot 2^x-1=0$$
Now, solving the quadratic equation in terms of $2^x$ as follows
$$2^x=frac{-3pmsqrt{(3)^2-4(1)(-1)}}{2(1)}=frac{-3pmsqrt{13}}{2}$$
But $2^x>0 forall xin mathbb{R}$ hence $$2^x=frac{-3+sqrt{13}}{2}$$
$$ln 2^x=lnleft(frac{-3+sqrt{13}}{2}right)$$
$$xln 2=lnleft(frac{sqrt{13}-3}{2}right)$$
$$x=color{red}{frac{lnleft(frac{sqrt{13}-3}{2}right)}{ln(2)}}$$
answered Jan 11 '16 at 14:46
Harish Chandra RajpootHarish Chandra Rajpoot
29.5k103671
$endgroup$
add a comment |
$begingroup$
you should be knowing $0.5^x=left(frac 12right)^{x}=frac 1{2^x}$, hence
$$0.5^{x}=2^x+3$$
$$frac 1{2^x}=2^x+3$$
$$(2^{x})^2+3cdot 2^x-1=0$$
Now, solving the quadratic equation in terms of $2^x$ as follows
$$2^x=frac{-3pmsqrt{(3)^2-4(1)(-1)}}{2(1)}=frac{-3pmsqrt{13}}{2}$$
But $2^x>0 forall xin mathbb{R}$ hence $$2^x=frac{-3+sqrt{13}}{2}$$
$$ln 2^x=lnleft(frac{-3+sqrt{13}}{2}right)$$
$$xln 2=lnleft(frac{sqrt{13}-3}{2}right)$$
$$x=color{red}{frac{lnleft(frac{sqrt{13}-3}{2}right)}{ln(2)}}$$
answered Jan 11 '16 at 14:46
Harish Chandra RajpootHarish Chandra Rajpoot
29.5k103671
$endgroup$
you should be knowing $0.5^x=left(frac 12right)^{x}=frac 1{2^x}$, hence
$$0.5^{x}=2^x+3$$
$$frac 1{2^x}=2^x+3$$
$$(2^{x})^2+3cdot 2^x-1=0$$
Now, solving the quadratic equation in terms of $2^x$ as follows
$$2^x=frac{-3pmsqrt{(3)^2-4(1)(-1)}}{2(1)}=frac{-3pmsqrt{13}}{2}$$
But $2^x>0 forall xin mathbb{R}$ hence $$2^x=frac{-3+sqrt{13}}{2}$$
$$ln 2^x=lnleft(frac{-3+sqrt{13}}{2}right)$$
$$xln 2=lnleft(frac{sqrt{13}-3}{2}right)$$
$$x=color{red}{frac{lnleft(frac{sqrt{13}-3}{2}right)}{ln(2)}}$$
answered Jan 11 '16 at 14:46
Harish Chandra RajpootHarish Chandra Rajpoot
29.5k103671
answered Jan 11 '16 at 14:46
Harish Chandra RajpootHarish Chandra Rajpoot
29.5k103671
answered Jan 11 '16 at 14:46
Harish Chandra RajpootHarish Chandra Rajpoot
29.5k103671
answered Jan 11 '16 at 14:46
Harish Chandra RajpootHarish Chandra Rajpoot
29.5k103671
29.5k103671
add a comment |
add a comment |
$begingroup$
$0.5^x=2^x +3$
$(frac{1}{2})^{x}=2^x + 3$
$frac{1}{2^{x}}=2^x + 3 $ $/2^x$
$1= 2^2x + 3* 2^x$
Now you make substitution, in these exercises method that is often used.
$y=2^x$
Our eqation looks like this:
$1=y^2 +3y$
$y^2 +3y-1=0$
$y=frac{-3pm sqrt{9+4}}{2}$
$y=frac{-3pm sqrt{13}}{2}$
$y_{1}=frac{-3-sqrt{13}}{2}$ or $ y_{2}=frac{-3+ sqrt{13}}{2}$
As we defined y= 2x, we know that it can not be negative number,so y1 is incorrect, and we continue only with y2.
$y=2^x$
$2^x=frac{-3+ sqrt{13}}{2} /log_{2}$
(log2(2x)=x)
$x=log_{2}frac{-3+sqrt{13}}{2}$
put it in calculator and you get :
x= -1.723
answered Jan 11 '16 at 14:43
florenceflorence
264
$endgroup$
$begingroup$
Welcome to Math Stackexchange! Did you know that you can type your math by surrounding it by$symbols?
$endgroup$
– JnxF
Jan 11 '16 at 14:56
add a comment |
$begingroup$
$0.5^x=2^x +3$
$(frac{1}{2})^{x}=2^x + 3$
$frac{1}{2^{x}}=2^x + 3 $ $/2^x$
$1= 2^2x + 3* 2^x$
Now you make substitution, in these exercises method that is often used.
$y=2^x$
Our eqation looks like this:
$1=y^2 +3y$
$y^2 +3y-1=0$
$y=frac{-3pm sqrt{9+4}}{2}$
$y=frac{-3pm sqrt{13}}{2}$
$y_{1}=frac{-3-sqrt{13}}{2}$ or $ y_{2}=frac{-3+ sqrt{13}}{2}$
As we defined y= 2x, we know that it can not be negative number,so y1 is incorrect, and we continue only with y2.
$y=2^x$
$2^x=frac{-3+ sqrt{13}}{2} /log_{2}$
(log2(2x)=x)
$x=log_{2}frac{-3+sqrt{13}}{2}$
put it in calculator and you get :
x= -1.723
answered Jan 11 '16 at 14:43
florenceflorence
264
$endgroup$
$begingroup$
Welcome to Math Stackexchange! Did you know that you can type your math by surrounding it by$symbols?
$endgroup$
– JnxF
Jan 11 '16 at 14:56
add a comment |
$begingroup$
$0.5^x=2^x +3$
$(frac{1}{2})^{x}=2^x + 3$
$frac{1}{2^{x}}=2^x + 3 $ $/2^x$
$1= 2^2x + 3* 2^x$
Now you make substitution, in these exercises method that is often used.
$y=2^x$
Our eqation looks like this:
$1=y^2 +3y$
$y^2 +3y-1=0$
$y=frac{-3pm sqrt{9+4}}{2}$
$y=frac{-3pm sqrt{13}}{2}$
$y_{1}=frac{-3-sqrt{13}}{2}$ or $ y_{2}=frac{-3+ sqrt{13}}{2}$
As we defined y= 2x, we know that it can not be negative number,so y1 is incorrect, and we continue only with y2.
$y=2^x$
$2^x=frac{-3+ sqrt{13}}{2} /log_{2}$
(log2(2x)=x)
$x=log_{2}frac{-3+sqrt{13}}{2}$
put it in calculator and you get :
x= -1.723
answered Jan 11 '16 at 14:43
florenceflorence
264
$endgroup$
$0.5^x=2^x +3$
$(frac{1}{2})^{x}=2^x + 3$
$frac{1}{2^{x}}=2^x + 3 $ $/2^x$
$1= 2^2x + 3* 2^x$
Now you make substitution, in these exercises method that is often used.
$y=2^x$
Our eqation looks like this:
$1=y^2 +3y$
$y^2 +3y-1=0$
$y=frac{-3pm sqrt{9+4}}{2}$
$y=frac{-3pm sqrt{13}}{2}$
$y_{1}=frac{-3-sqrt{13}}{2}$ or $ y_{2}=frac{-3+ sqrt{13}}{2}$
As we defined y= 2x, we know that it can not be negative number,so y1 is incorrect, and we continue only with y2.
$y=2^x$
$2^x=frac{-3+ sqrt{13}}{2} /log_{2}$
(log2(2x)=x)
$x=log_{2}frac{-3+sqrt{13}}{2}$
put it in calculator and you get :
x= -1.723
answered Jan 11 '16 at 14:43
florenceflorence
264
edited Jan 11 '16 at 16:50
answered Jan 11 '16 at 14:43
florenceflorence
264
answered Jan 11 '16 at 14:43
florenceflorence
264
answered Jan 11 '16 at 14:43
florenceflorence
264
264
$begingroup$
Welcome to Math Stackexchange! Did you know that you can type your math by surrounding it by$symbols?
$endgroup$
– JnxF
Jan 11 '16 at 14:56
add a comment |
$begingroup$
Welcome to Math Stackexchange! Did you know that you can type your math by surrounding it by$symbols?
$endgroup$
– JnxF
Jan 11 '16 at 14:56
$begingroup$
Welcome to Math Stackexchange! Did you know that you can type your math by surrounding it by
$ symbols?$endgroup$
– JnxF
Jan 11 '16 at 14:56
$begingroup$
Welcome to Math Stackexchange! Did you know that you can type your math by surrounding it by
$ symbols?$endgroup$
– JnxF
Jan 11 '16 at 14:56
add a comment |
$begingroup$
This may be a approach to solve it, hope this helps.
Thank you!
answered Jan 7 at 6:47
user629353user629353
1187
$endgroup$
add a comment |
$begingroup$
This may be a approach to solve it, hope this helps.
Thank you!
answered Jan 7 at 6:47
user629353user629353
1187
$endgroup$
add a comment |
$begingroup$
This may be a approach to solve it, hope this helps.
Thank you!
answered Jan 7 at 6:47
user629353user629353
1187
$endgroup$
This may be a approach to solve it, hope this helps.
Thank you!
answered Jan 7 at 6:47
user629353user629353
1187
answered Jan 7 at 6:47
user629353user629353
1187
answered Jan 7 at 6:47
user629353user629353
1187
answered Jan 7 at 6:47
user629353user629353
1187
1187
add a comment |
add a comment |
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