How to find $x$ in $log_9 (x)+log_9 (x-2)=log_9 (8)$
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How would I go about solving this problem? I know that you can remove the $log$s if the have the same base resulting leaving $x(x-2)=8$, but what would be the next step to finding the value of $x$?
logarithms
edited Sep 6 at 7:01
GoodDeeds
10.2k21335
asked Sep 6 at 6:26
Confused Student
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up vote
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How would I go about solving this problem? I know that you can remove the $log$s if the have the same base resulting leaving $x(x-2)=8$, but what would be the next step to finding the value of $x$?
logarithms
edited Sep 6 at 7:01
GoodDeeds
10.2k21335
asked Sep 6 at 6:26
Confused Student
92
1
Do you mean $log_9$ not $log9$?
– TheSimpliFire
Sep 6 at 6:29
The equation is just a quadratic! You have $$x^2-2x-8=(x-4)(x+2)=0$$ and just remember that $x>2$.
– TheSimpliFire
Sep 6 at 6:30
I think the exercise meant $log_9$ instead of $log 9$, doesn't seems like you can go on "adding basis to the logarithm" as you said. If that's the case you can proceed like those answers below.
– Robson
Sep 6 at 6:37
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
How would I go about solving this problem? I know that you can remove the $log$s if the have the same base resulting leaving $x(x-2)=8$, but what would be the next step to finding the value of $x$?
logarithms
edited Sep 6 at 7:01
GoodDeeds
10.2k21335
asked Sep 6 at 6:26
Confused Student
92
How would I go about solving this problem? I know that you can remove the $log$s if the have the same base resulting leaving $x(x-2)=8$, but what would be the next step to finding the value of $x$?
logarithms
logarithms
edited Sep 6 at 7:01
GoodDeeds
10.2k21335
asked Sep 6 at 6:26
Confused Student
92
edited Sep 6 at 7:01
GoodDeeds
10.2k21335
asked Sep 6 at 6:26
Confused Student
92
edited Sep 6 at 7:01
GoodDeeds
10.2k21335
edited Sep 6 at 7:01
GoodDeeds
10.2k21335
edited Sep 6 at 7:01
GoodDeeds
10.2k21335
10.2k21335
asked Sep 6 at 6:26
Confused Student
92
asked Sep 6 at 6:26
Confused Student
92
asked Sep 6 at 6:26
Confused Student
92
92
1
Do you mean $log_9$ not $log9$?
– TheSimpliFire
Sep 6 at 6:29
The equation is just a quadratic! You have $$x^2-2x-8=(x-4)(x+2)=0$$ and just remember that $x>2$.
– TheSimpliFire
Sep 6 at 6:30
I think the exercise meant $log_9$ instead of $log 9$, doesn't seems like you can go on "adding basis to the logarithm" as you said. If that's the case you can proceed like those answers below.
– Robson
Sep 6 at 6:37
add a comment |Â
1
Do you mean $log_9$ not $log9$?
– TheSimpliFire
Sep 6 at 6:29
The equation is just a quadratic! You have $$x^2-2x-8=(x-4)(x+2)=0$$ and just remember that $x>2$.
– TheSimpliFire
Sep 6 at 6:30
I think the exercise meant $log_9$ instead of $log 9$, doesn't seems like you can go on "adding basis to the logarithm" as you said. If that's the case you can proceed like those answers below.
– Robson
Sep 6 at 6:37
1
1
Do you mean $log_9$ not $log9$?
– TheSimpliFire
Sep 6 at 6:29
Do you mean $log_9$ not $log9$?
– TheSimpliFire
Sep 6 at 6:29
The equation is just a quadratic! You have $$x^2-2x-8=(x-4)(x+2)=0$$ and just remember that $x>2$.
– TheSimpliFire
Sep 6 at 6:30
The equation is just a quadratic! You have $$x^2-2x-8=(x-4)(x+2)=0$$ and just remember that $x>2$.
– TheSimpliFire
Sep 6 at 6:30
I think the exercise meant $log_9$ instead of $log 9$, doesn't seems like you can go on "adding basis to the logarithm" as you said. If that's the case you can proceed like those answers below.
– Robson
Sep 6 at 6:37
I think the exercise meant $log_9$ instead of $log 9$, doesn't seems like you can go on "adding basis to the logarithm" as you said. If that's the case you can proceed like those answers below.
– Robson
Sep 6 at 6:37
add a comment |Â
2 Answers
2
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up vote
1
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Hint: $x(x-2) = 8$ if we distribute, $x^2 - 2x = 8$, or $x^2-2x-8 = 0$. Can you proceed from here?
answered Sep 6 at 6:30
rb612
1,754822
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up vote
1
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$$x(x-2)=8$$
$$x^2-2x=8$$
$$x^2-2x-8=0$$
$$(x-4)(x+2)=0$$
But as you can see we want $x>0 $ and $x-2>0$ because of the domain of $log x $ is $x>0$
Thus $x>2$
Thus $x=4$ is the only solution to this equation
answered Sep 6 at 6:30
Deepesh Meena
3,8912825
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint: $x(x-2) = 8$ if we distribute, $x^2 - 2x = 8$, or $x^2-2x-8 = 0$. Can you proceed from here?
answered Sep 6 at 6:30
rb612
1,754822
add a comment |Â
up vote
1
down vote
Hint: $x(x-2) = 8$ if we distribute, $x^2 - 2x = 8$, or $x^2-2x-8 = 0$. Can you proceed from here?
answered Sep 6 at 6:30
rb612
1,754822
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: $x(x-2) = 8$ if we distribute, $x^2 - 2x = 8$, or $x^2-2x-8 = 0$. Can you proceed from here?
answered Sep 6 at 6:30
rb612
1,754822
Hint: $x(x-2) = 8$ if we distribute, $x^2 - 2x = 8$, or $x^2-2x-8 = 0$. Can you proceed from here?
answered Sep 6 at 6:30
rb612
1,754822
answered Sep 6 at 6:30
rb612
1,754822
answered Sep 6 at 6:30
rb612
1,754822
answered Sep 6 at 6:30
rb612
1,754822
1,754822
add a comment |Â
add a comment |Â
up vote
1
down vote
$$x(x-2)=8$$
$$x^2-2x=8$$
$$x^2-2x-8=0$$
$$(x-4)(x+2)=0$$
But as you can see we want $x>0 $ and $x-2>0$ because of the domain of $log x $ is $x>0$
Thus $x>2$
Thus $x=4$ is the only solution to this equation
answered Sep 6 at 6:30
Deepesh Meena
3,8912825
add a comment |Â
up vote
1
down vote
$$x(x-2)=8$$
$$x^2-2x=8$$
$$x^2-2x-8=0$$
$$(x-4)(x+2)=0$$
But as you can see we want $x>0 $ and $x-2>0$ because of the domain of $log x $ is $x>0$
Thus $x>2$
Thus $x=4$ is the only solution to this equation
answered Sep 6 at 6:30
Deepesh Meena
3,8912825
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$x(x-2)=8$$
$$x^2-2x=8$$
$$x^2-2x-8=0$$
$$(x-4)(x+2)=0$$
But as you can see we want $x>0 $ and $x-2>0$ because of the domain of $log x $ is $x>0$
Thus $x>2$
Thus $x=4$ is the only solution to this equation
answered Sep 6 at 6:30
Deepesh Meena
3,8912825
$$x(x-2)=8$$
$$x^2-2x=8$$
$$x^2-2x-8=0$$
$$(x-4)(x+2)=0$$
But as you can see we want $x>0 $ and $x-2>0$ because of the domain of $log x $ is $x>0$
Thus $x>2$
Thus $x=4$ is the only solution to this equation
answered Sep 6 at 6:30
Deepesh Meena
3,8912825
answered Sep 6 at 6:30
Deepesh Meena
3,8912825
answered Sep 6 at 6:30
Deepesh Meena
3,8912825
answered Sep 6 at 6:30
Deepesh Meena
3,8912825
3,8912825
add a comment |Â
add a comment |Â
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1
Do you mean $log_9$ not $log9$?
– TheSimpliFire
Sep 6 at 6:29
The equation is just a quadratic! You have $$x^2-2x-8=(x-4)(x+2)=0$$ and just remember that $x>2$.
– TheSimpliFire
Sep 6 at 6:30
I think the exercise meant $log_9$ instead of $log 9$, doesn't seems like you can go on "adding basis to the logarithm" as you said. If that's the case you can proceed like those answers below.
– Robson
Sep 6 at 6:37