Help find my computational error for logarithms
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I am supposed to get $x = 8$ and $x = x^{-2/3}$. What did I do wrong?
algebra-precalculus logarithms
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add a comment |
$begingroup$
I am supposed to get $x = 8$ and $x = x^{-2/3}$. What did I do wrong?
algebra-precalculus logarithms
$endgroup$
$begingroup$
$frac 1 {log x}$ is not generally the same as $log frac 1 x $
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– J. W. Tanner
52 mins ago
$begingroup$
Did you mean $x=mathbf 2^{-2/3}$ ?
$endgroup$
– J. W. Tanner
37 mins ago
add a comment |
$begingroup$
I am supposed to get $x = 8$ and $x = x^{-2/3}$. What did I do wrong?
algebra-precalculus logarithms
$endgroup$
I am supposed to get $x = 8$ and $x = x^{-2/3}$. What did I do wrong?
algebra-precalculus logarithms
algebra-precalculus logarithms
edited 50 mins ago
Eevee Trainer
7,72621338
7,72621338
asked 56 mins ago
KevinKevin
446
446
$begingroup$
$frac 1 {log x}$ is not generally the same as $log frac 1 x $
$endgroup$
– J. W. Tanner
52 mins ago
$begingroup$
Did you mean $x=mathbf 2^{-2/3}$ ?
$endgroup$
– J. W. Tanner
37 mins ago
add a comment |
$begingroup$
$frac 1 {log x}$ is not generally the same as $log frac 1 x $
$endgroup$
– J. W. Tanner
52 mins ago
$begingroup$
Did you mean $x=mathbf 2^{-2/3}$ ?
$endgroup$
– J. W. Tanner
37 mins ago
$begingroup$
$frac 1 {log x}$ is not generally the same as $log frac 1 x $
$endgroup$
– J. W. Tanner
52 mins ago
$begingroup$
$frac 1 {log x}$ is not generally the same as $log frac 1 x $
$endgroup$
– J. W. Tanner
52 mins ago
$begingroup$
Did you mean $x=mathbf 2^{-2/3}$ ?
$endgroup$
– J. W. Tanner
37 mins ago
$begingroup$
Did you mean $x=mathbf 2^{-2/3}$ ?
$endgroup$
– J. W. Tanner
37 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that
$$left( log_2(x) right)^{-1} neq log_2 left(x^{-1} right)$$
For example, take $x = 4$. Then
$$left( log_2(x) right)^{-1} = left( log_2(4) right)^{-1} = 2^{-1} = frac 1 2$$
but
$$log_2 left(x^{-1} right) = log_2 left( frac 1 4 right) = -2$$
This is where your error lies.
$endgroup$
add a comment |
$begingroup$
To get the correct answer, let $L=log_2(x).$
Then we have $$frac 1 2 L - frac 1 L = frac 7 6.$$
Multiply by $6L$ to get $$3L^2-6=7L.$$
Thus $$3L^2-7L-6=0$$
or $$(3L+2)(L-3)=0.$$
Can you take it from here?
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1
$begingroup$
Not gonna lie, that's a pretty slick solution (even if it's just a minor trick involving a substitution - I always love it when something like that makes the work look 10 times easier :p).
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– Eevee Trainer
41 mins ago
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I know how to solve in this way, I was just confused about why I got a different answer when using another approach. But still thank you for your help ;)
$endgroup$
– Kevin
26 mins ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that
$$left( log_2(x) right)^{-1} neq log_2 left(x^{-1} right)$$
For example, take $x = 4$. Then
$$left( log_2(x) right)^{-1} = left( log_2(4) right)^{-1} = 2^{-1} = frac 1 2$$
but
$$log_2 left(x^{-1} right) = log_2 left( frac 1 4 right) = -2$$
This is where your error lies.
$endgroup$
add a comment |
$begingroup$
Note that
$$left( log_2(x) right)^{-1} neq log_2 left(x^{-1} right)$$
For example, take $x = 4$. Then
$$left( log_2(x) right)^{-1} = left( log_2(4) right)^{-1} = 2^{-1} = frac 1 2$$
but
$$log_2 left(x^{-1} right) = log_2 left( frac 1 4 right) = -2$$
This is where your error lies.
$endgroup$
add a comment |
$begingroup$
Note that
$$left( log_2(x) right)^{-1} neq log_2 left(x^{-1} right)$$
For example, take $x = 4$. Then
$$left( log_2(x) right)^{-1} = left( log_2(4) right)^{-1} = 2^{-1} = frac 1 2$$
but
$$log_2 left(x^{-1} right) = log_2 left( frac 1 4 right) = -2$$
This is where your error lies.
$endgroup$
Note that
$$left( log_2(x) right)^{-1} neq log_2 left(x^{-1} right)$$
For example, take $x = 4$. Then
$$left( log_2(x) right)^{-1} = left( log_2(4) right)^{-1} = 2^{-1} = frac 1 2$$
but
$$log_2 left(x^{-1} right) = log_2 left( frac 1 4 right) = -2$$
This is where your error lies.
answered 53 mins ago
Eevee TrainerEevee Trainer
7,72621338
7,72621338
add a comment |
add a comment |
$begingroup$
To get the correct answer, let $L=log_2(x).$
Then we have $$frac 1 2 L - frac 1 L = frac 7 6.$$
Multiply by $6L$ to get $$3L^2-6=7L.$$
Thus $$3L^2-7L-6=0$$
or $$(3L+2)(L-3)=0.$$
Can you take it from here?
$endgroup$
1
$begingroup$
Not gonna lie, that's a pretty slick solution (even if it's just a minor trick involving a substitution - I always love it when something like that makes the work look 10 times easier :p).
$endgroup$
– Eevee Trainer
41 mins ago
$begingroup$
I know how to solve in this way, I was just confused about why I got a different answer when using another approach. But still thank you for your help ;)
$endgroup$
– Kevin
26 mins ago
add a comment |
$begingroup$
To get the correct answer, let $L=log_2(x).$
Then we have $$frac 1 2 L - frac 1 L = frac 7 6.$$
Multiply by $6L$ to get $$3L^2-6=7L.$$
Thus $$3L^2-7L-6=0$$
or $$(3L+2)(L-3)=0.$$
Can you take it from here?
$endgroup$
1
$begingroup$
Not gonna lie, that's a pretty slick solution (even if it's just a minor trick involving a substitution - I always love it when something like that makes the work look 10 times easier :p).
$endgroup$
– Eevee Trainer
41 mins ago
$begingroup$
I know how to solve in this way, I was just confused about why I got a different answer when using another approach. But still thank you for your help ;)
$endgroup$
– Kevin
26 mins ago
add a comment |
$begingroup$
To get the correct answer, let $L=log_2(x).$
Then we have $$frac 1 2 L - frac 1 L = frac 7 6.$$
Multiply by $6L$ to get $$3L^2-6=7L.$$
Thus $$3L^2-7L-6=0$$
or $$(3L+2)(L-3)=0.$$
Can you take it from here?
$endgroup$
To get the correct answer, let $L=log_2(x).$
Then we have $$frac 1 2 L - frac 1 L = frac 7 6.$$
Multiply by $6L$ to get $$3L^2-6=7L.$$
Thus $$3L^2-7L-6=0$$
or $$(3L+2)(L-3)=0.$$
Can you take it from here?
answered 47 mins ago
J. W. TannerJ. W. Tanner
3,0701320
3,0701320
1
$begingroup$
Not gonna lie, that's a pretty slick solution (even if it's just a minor trick involving a substitution - I always love it when something like that makes the work look 10 times easier :p).
$endgroup$
– Eevee Trainer
41 mins ago
$begingroup$
I know how to solve in this way, I was just confused about why I got a different answer when using another approach. But still thank you for your help ;)
$endgroup$
– Kevin
26 mins ago
add a comment |
1
$begingroup$
Not gonna lie, that's a pretty slick solution (even if it's just a minor trick involving a substitution - I always love it when something like that makes the work look 10 times easier :p).
$endgroup$
– Eevee Trainer
41 mins ago
$begingroup$
I know how to solve in this way, I was just confused about why I got a different answer when using another approach. But still thank you for your help ;)
$endgroup$
– Kevin
26 mins ago
1
1
$begingroup$
Not gonna lie, that's a pretty slick solution (even if it's just a minor trick involving a substitution - I always love it when something like that makes the work look 10 times easier :p).
$endgroup$
– Eevee Trainer
41 mins ago
$begingroup$
Not gonna lie, that's a pretty slick solution (even if it's just a minor trick involving a substitution - I always love it when something like that makes the work look 10 times easier :p).
$endgroup$
– Eevee Trainer
41 mins ago
$begingroup$
I know how to solve in this way, I was just confused about why I got a different answer when using another approach. But still thank you for your help ;)
$endgroup$
– Kevin
26 mins ago
$begingroup$
I know how to solve in this way, I was just confused about why I got a different answer when using another approach. But still thank you for your help ;)
$endgroup$
– Kevin
26 mins ago
add a comment |
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$begingroup$
$frac 1 {log x}$ is not generally the same as $log frac 1 x $
$endgroup$
– J. W. Tanner
52 mins ago
$begingroup$
Did you mean $x=mathbf 2^{-2/3}$ ?
$endgroup$
– J. W. Tanner
37 mins ago