Solving $9^{1+log x} - 3^{1+log x} - 210 = 0$ where base of log is $3$ for $x$












5












$begingroup$


Question is to solve the equation for value of $x$.




$$9^{1+log x} - 3^{1+log x} - 210 = 0; quad text{where base of log is }3$$




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]






logarithms







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  • $begingroup$
    Use $a^{log_b(x)} = x^{log_b(a)}$.
    $endgroup$
    – Viktor Glombik
    5 hours 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
    5 hours ago


















5












$begingroup$


Question is to solve the equation for value of $x$.




$$9^{1+log x} - 3^{1+log x} - 210 = 0; quad text{where base of log is }3$$




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]






logarithms







share|cite|improve this question









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
    5 hours 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
    5 hours ago
















5












5








5





$begingroup$


Question is to solve the equation for value of $x$.




$$9^{1+log x} - 3^{1+log x} - 210 = 0; quad text{where base of log is }3$$




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]






logarithms







share|cite|improve this question









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 $x$.




$$9^{1+log x} - 3^{1+log x} - 210 = 0; quad text{where base of log is }3$$




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]






logarithms




logarithms






share|cite|improve this question









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.











share|cite|improve this question









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.









share|cite|improve this question




share|cite|improve this question








edited 4 hours ago









Asaf Karagila♦

309k33441775




309k33441775






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 5 hours 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.
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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
    5 hours 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
    5 hours ago




















  • $begingroup$
    Use $a^{log_b(x)} = x^{log_b(a)}$.
    $endgroup$
    – Viktor Glombik
    5 hours 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
    5 hours ago


















$begingroup$
Use $a^{log_b(x)} = x^{log_b(a)}$.
$endgroup$
– Viktor Glombik
5 hours ago




$begingroup$
Use $a^{log_b(x)} = x^{log_b(a)}$.
$endgroup$
– Viktor Glombik
5 hours 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
5 hours 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
5 hours ago












1 Answer
1






active

oldest

votes


















5












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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
    5 hours ago










  • $begingroup$
    He meant$(3x+14)(x-5) = 0$. He was solving a quadratic. Must be $times$ in the middle.
    $endgroup$
    – Vizag
    5 hours ago










  • $begingroup$
    😅😅Oh! I didn't noticed that.
    $endgroup$
    – Piyush Raj
    4 hours ago












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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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
    5 hours ago










  • $begingroup$
    He meant$(3x+14)(x-5) = 0$. He was solving a quadratic. Must be $times$ in the middle.
    $endgroup$
    – Vizag
    5 hours ago










  • $begingroup$
    😅😅Oh! I didn't noticed that.
    $endgroup$
    – Piyush Raj
    4 hours ago
















5












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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
    5 hours ago










  • $begingroup$
    He meant$(3x+14)(x-5) = 0$. He was solving a quadratic. Must be $times$ in the middle.
    $endgroup$
    – Vizag
    5 hours ago










  • $begingroup$
    😅😅Oh! I didn't noticed that.
    $endgroup$
    – Piyush Raj
    4 hours ago














5












5








5





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 5 hours ago









DMcMor

2,91321328




2,91321328










answered 5 hours ago









VizagVizag

534113




534113








  • 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
    5 hours ago










  • $begingroup$
    He meant$(3x+14)(x-5) = 0$. He was solving a quadratic. Must be $times$ in the middle.
    $endgroup$
    – Vizag
    5 hours ago










  • $begingroup$
    😅😅Oh! I didn't noticed that.
    $endgroup$
    – Piyush Raj
    4 hours ago














  • 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
    5 hours ago










  • $begingroup$
    He meant$(3x+14)(x-5) = 0$. He was solving a quadratic. Must be $times$ in the middle.
    $endgroup$
    – Vizag
    5 hours ago










  • $begingroup$
    😅😅Oh! I didn't noticed that.
    $endgroup$
    – Piyush Raj
    4 hours 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
5 hours 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
5 hours ago












$begingroup$
He meant$(3x+14)(x-5) = 0$. He was solving a quadratic. Must be $times$ in the middle.
$endgroup$
– Vizag
5 hours ago




$begingroup$
He meant$(3x+14)(x-5) = 0$. He was solving a quadratic. Must be $times$ in the middle.
$endgroup$
– Vizag
5 hours ago












$begingroup$
😅😅Oh! I didn't noticed that.
$endgroup$
– Piyush Raj
4 hours ago




$begingroup$
😅😅Oh! I didn't noticed that.
$endgroup$
– Piyush Raj
4 hours ago










Piyush Raj is a new contributor. Be nice, and check out our Code of Conduct.










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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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