Solve equation for value of x: Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Interpret a linear scale as a logarithmic scaleHow to solve logarithm word problem given the exponential equation?How to solve this equation algebraically?What went wrong in these solutions of $log big(x^log xbig)=4$Trouble solving for exponents with constantsWhat is the value of $x$ in this equation using logarithmsWhat is solution of this logarithmic equationThe solution of the equation $7^x+7 =8^x$ can be expressed in form $x=log7^7$ to the base $b$. What is $b$?how to solve the logarithmic equation which has both n and lognSolve the equation for x:
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Solve equation for value of x:
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Interpret a linear scale as a logarithmic scaleHow to solve logarithm word problem given the exponential equation?How to solve this equation algebraically?What went wrong in these solutions of $log big(x^log xbig)=4$Trouble solving for exponents with constantsWhat is the value of $x$ in this equation using logarithmsWhat is solution of this logarithmic equationThe solution of the equation $7^x+7 =8^x$ can be expressed in form $x=log7^7$ to the base $b$. What is $b$?how to solve the logarithmic equation which has both n and lognSolve the equation for x:
$begingroup$
Question is to solve the equation for value of $x$.
$$9^1+log x - 3^1+log x - 210 = 0; quad textwhere base of log is 3$$
The answer given is $x=5$
I've tried to solve it. And got two values of $x= -14/3$ and $x=5$. What I've done wrong?
![[enter image description here]](https://i.stack.imgur.com/OR8Rm.jpg)
logarithms
edited 53 mins ago
John Doe
12.1k11340
asked 57 mins ago
Piyush RajPiyush Raj
304
New contributor
Piyush Raj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Question is to solve the equation for value of $x$.
$$9^1+log x - 3^1+log x - 210 = 0; quad textwhere base of log is 3$$
The answer given is $x=5$
I've tried to solve it. And got two values of $x= -14/3$ and $x=5$. What I've done wrong?
logarithms
edited 53 mins ago
John Doe
12.1k11340
asked 57 mins ago
Piyush RajPiyush Raj
304
New contributor
Piyush Raj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Use $a^log_b(x) = x^log_b(a)$.
$endgroup$
– Viktor Glombik
50 mins ago
1
$begingroup$
The final step should be $(3k+14)(k-5)=0$ (multiplication since you are factoring) This is what then allows you to reach your conclusion.
$endgroup$
– John Doe
47 mins ago
add a comment |
$begingroup$
Question is to solve the equation for value of $x$.
$$9^1+log x - 3^1+log x - 210 = 0; quad textwhere base of log is 3$$
The answer given is $x=5$
I've tried to solve it. And got two values of $x= -14/3$ and $x=5$. What I've done wrong?
logarithms
edited 53 mins ago
John Doe
12.1k11340
asked 57 mins ago
Piyush RajPiyush Raj
304
New contributor
Piyush Raj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Question is to solve the equation for value of $x$.
$$9^1+log x - 3^1+log x - 210 = 0; quad textwhere base of log is 3$$
The answer given is $x=5$
I've tried to solve it. And got two values of $x= -14/3$ and $x=5$. What I've done wrong?
logarithms
logarithms
edited 53 mins ago
John Doe
12.1k11340
asked 57 mins ago
Piyush RajPiyush Raj
304
New contributor
Piyush Raj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 53 mins ago
John Doe
12.1k11340
asked 57 mins ago
Piyush RajPiyush Raj
304
New contributor
Piyush Raj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 53 mins ago
John Doe
12.1k11340
edited 53 mins ago
John Doe
12.1k11340
edited 53 mins ago
John Doe
12.1k11340
12.1k11340
asked 57 mins ago
Piyush RajPiyush Raj
304
New contributor
Piyush Raj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 57 mins ago
Piyush RajPiyush Raj
304
asked 57 mins ago
Piyush RajPiyush Raj
304
304
New contributor
Piyush Raj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Piyush Raj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Piyush Raj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Use $a^log_b(x) = x^log_b(a)$.
$endgroup$
– Viktor Glombik
50 mins ago
1
$begingroup$
The final step should be $(3k+14)(k-5)=0$ (multiplication since you are factoring) This is what then allows you to reach your conclusion.
$endgroup$
– John Doe
47 mins ago
add a comment |
$begingroup$
Use $a^log_b(x) = x^log_b(a)$.
$endgroup$
– Viktor Glombik
50 mins ago
1
$begingroup$
The final step should be $(3k+14)(k-5)=0$ (multiplication since you are factoring) This is what then allows you to reach your conclusion.
$endgroup$
– John Doe
47 mins ago
$begingroup$
Use $a^log_b(x) = x^log_b(a)$.
$endgroup$
– Viktor Glombik
50 mins ago
$begingroup$
Use $a^log_b(x) = x^log_b(a)$.
$endgroup$
– Viktor Glombik
50 mins ago
1
1
$begingroup$
The final step should be $(3k+14)(k-5)=0$ (multiplication since you are factoring) This is what then allows you to reach your conclusion.
$endgroup$
– John Doe
47 mins ago
$begingroup$
The final step should be $(3k+14)(k-5)=0$ (multiplication since you are factoring) This is what then allows you to reach your conclusion.
$endgroup$
– John Doe
47 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have solved correctly just made one error towards the end.
Note that the domain of $log(x)$ is $x > 0$ so $x=-14/3$ is rejected as it is not in the domain of the function.
edited 50 mins ago
DMcMor
2,91321328
answered 52 mins ago
VizagVizag
514112
$endgroup$
2
$begingroup$
Yep. And going from $(3 x + 14) + (x - 5) = 0$ does not mean that either term in the sum is $0$.
$endgroup$
– David G. Stork
49 mins ago
$begingroup$
He meant$(3x+14)(x-5) = 0$. He was solving a quadratic. Must be $times$ in the middle.
$endgroup$
– Vizag
47 mins ago
$begingroup$
😅😅Oh! I didn't noticed that.
$endgroup$
– Piyush Raj
13 mins ago
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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votes
$begingroup$
You have solved correctly just made one error towards the end.
Note that the domain of $log(x)$ is $x > 0$ so $x=-14/3$ is rejected as it is not in the domain of the function.
edited 50 mins ago
DMcMor
2,91321328
answered 52 mins ago
VizagVizag
514112
$endgroup$
2
$begingroup$
Yep. And going from $(3 x + 14) + (x - 5) = 0$ does not mean that either term in the sum is $0$.
$endgroup$
– David G. Stork
49 mins ago
$begingroup$
He meant$(3x+14)(x-5) = 0$. He was solving a quadratic. Must be $times$ in the middle.
$endgroup$
– Vizag
47 mins ago
$begingroup$
😅😅Oh! I didn't noticed that.
$endgroup$
– Piyush Raj
13 mins ago
add a comment |
$begingroup$
You have solved correctly just made one error towards the end.
Note that the domain of $log(x)$ is $x > 0$ so $x=-14/3$ is rejected as it is not in the domain of the function.
edited 50 mins ago
DMcMor
2,91321328
answered 52 mins ago
VizagVizag
514112
$endgroup$
2
$begingroup$
Yep. And going from $(3 x + 14) + (x - 5) = 0$ does not mean that either term in the sum is $0$.
$endgroup$
– David G. Stork
49 mins ago
$begingroup$
He meant$(3x+14)(x-5) = 0$. He was solving a quadratic. Must be $times$ in the middle.
$endgroup$
– Vizag
47 mins ago
$begingroup$
😅😅Oh! I didn't noticed that.
$endgroup$
– Piyush Raj
13 mins ago
add a comment |
$begingroup$
You have solved correctly just made one error towards the end.
Note that the domain of $log(x)$ is $x > 0$ so $x=-14/3$ is rejected as it is not in the domain of the function.
edited 50 mins ago
DMcMor
2,91321328
answered 52 mins ago
VizagVizag
514112
$endgroup$
You have solved correctly just made one error towards the end.
Note that the domain of $log(x)$ is $x > 0$ so $x=-14/3$ is rejected as it is not in the domain of the function.
edited 50 mins ago
DMcMor
2,91321328
answered 52 mins ago
VizagVizag
514112
edited 50 mins ago
DMcMor
2,91321328
edited 50 mins ago
DMcMor
2,91321328
edited 50 mins ago
DMcMor
2,91321328
2,91321328
answered 52 mins ago
VizagVizag
514112
answered 52 mins ago
VizagVizag
514112
answered 52 mins ago
VizagVizag
514112
514112
2
$begingroup$
Yep. And going from $(3 x + 14) + (x - 5) = 0$ does not mean that either term in the sum is $0$.
$endgroup$
– David G. Stork
49 mins ago
$begingroup$
He meant$(3x+14)(x-5) = 0$. He was solving a quadratic. Must be $times$ in the middle.
$endgroup$
– Vizag
47 mins ago
$begingroup$
😅😅Oh! I didn't noticed that.
$endgroup$
– Piyush Raj
13 mins ago
add a comment |
2
$begingroup$
Yep. And going from $(3 x + 14) + (x - 5) = 0$ does not mean that either term in the sum is $0$.
$endgroup$
– David G. Stork
49 mins ago
$begingroup$
He meant$(3x+14)(x-5) = 0$. He was solving a quadratic. Must be $times$ in the middle.
$endgroup$
– Vizag
47 mins ago
$begingroup$
😅😅Oh! I didn't noticed that.
$endgroup$
– Piyush Raj
13 mins ago
2
2
$begingroup$
Yep. And going from $(3 x + 14) + (x - 5) = 0$ does not mean that either term in the sum is $0$.
$endgroup$
– David G. Stork
49 mins ago
$begingroup$
Yep. And going from $(3 x + 14) + (x - 5) = 0$ does not mean that either term in the sum is $0$.
$endgroup$
– David G. Stork
49 mins ago
$begingroup$
He meant$(3x+14)(x-5) = 0$. He was solving a quadratic. Must be $times$ in the middle.
$endgroup$
– Vizag
47 mins ago
$begingroup$
He meant$(3x+14)(x-5) = 0$. He was solving a quadratic. Must be $times$ in the middle.
$endgroup$
– Vizag
47 mins ago
$begingroup$
😅😅Oh! I didn't noticed that.
$endgroup$
– Piyush Raj
13 mins ago
$begingroup$
😅😅Oh! I didn't noticed that.
$endgroup$
– Piyush Raj
13 mins ago
add a comment |
Piyush Raj is a new contributor. Be nice, and check out our Code of Conduct.
Piyush Raj is a new contributor. Be nice, and check out our Code of Conduct.
Piyush Raj is a new contributor. Be nice, and check out our Code of Conduct.
Piyush Raj is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Use $a^log_b(x) = x^log_b(a)$.
$endgroup$
– Viktor Glombik
50 mins ago
1
$begingroup$
The final step should be $(3k+14)(k-5)=0$ (multiplication since you are factoring) This is what then allows you to reach your conclusion.
$endgroup$
– John Doe
47 mins ago