Help find my computational error for logarithmsWhat are logarithms, and what do they do?Difference of Logarithms to form a quotient?simple logarithms with exponentsSimple question on logarithmsSolving exponential equations using logarithmsI need help identifying the slope for an equation.Solutions of $2^x 7^1/xle 14$Adding logarithms with different basesSolving Logarithms HelpHow do you multiply these logarithms and find the domain?
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Help find my computational error for logarithms
What are logarithms, and what do they do?Difference of Logarithms to form a quotient?simple logarithms with exponentsSimple question on logarithmsSolving exponential equations using logarithmsI need help identifying the slope for an equation.Solutions of $2^x 7^1/xle 14$Adding logarithms with different basesSolving Logarithms HelpHow do you multiply these logarithms and find the domain?
$begingroup$
I am supposed to get $x = 8$ and $x = x^-2/3$. What did I do wrong?
algebra-precalculus logarithms
edited 53 mins ago
Eevee Trainer
7,72621338
asked 59 mins ago
KevinKevin
446
$endgroup$
add a comment |
$begingroup$
I am supposed to get $x = 8$ and $x = x^-2/3$. What did I do wrong?
algebra-precalculus logarithms
edited 53 mins ago
Eevee Trainer
7,72621338
asked 59 mins ago
KevinKevin
446
$endgroup$
$begingroup$
$frac 1 log x$ is not generally the same as $log frac 1 x $
$endgroup$
– J. W. Tanner
55 mins ago
$begingroup$
Did you mean $x=mathbf 2^-2/3$ ?
$endgroup$
– J. W. Tanner
40 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
edited 53 mins ago
Eevee Trainer
7,72621338
asked 59 mins ago
KevinKevin
446
$endgroup$
I am supposed to get $x = 8$ and $x = x^-2/3$. What did I do wrong?
algebra-precalculus logarithms
algebra-precalculus logarithms
edited 53 mins ago
Eevee Trainer
7,72621338
asked 59 mins ago
KevinKevin
446
edited 53 mins ago
Eevee Trainer
7,72621338
asked 59 mins ago
KevinKevin
446
edited 53 mins ago
Eevee Trainer
7,72621338
edited 53 mins ago
Eevee Trainer
7,72621338
edited 53 mins ago
Eevee Trainer
7,72621338
7,72621338
asked 59 mins ago
KevinKevin
446
asked 59 mins ago
KevinKevin
446
asked 59 mins ago
KevinKevin
446
446
$begingroup$
$frac 1 log x$ is not generally the same as $log frac 1 x $
$endgroup$
– J. W. Tanner
55 mins ago
$begingroup$
Did you mean $x=mathbf 2^-2/3$ ?
$endgroup$
– J. W. Tanner
40 mins ago
add a comment |
$begingroup$
$frac 1 log x$ is not generally the same as $log frac 1 x $
$endgroup$
– J. W. Tanner
55 mins ago
$begingroup$
Did you mean $x=mathbf 2^-2/3$ ?
$endgroup$
– J. W. Tanner
40 mins ago
$begingroup$
$frac 1 log x$ is not generally the same as $log frac 1 x $
$endgroup$
– J. W. Tanner
55 mins ago
$begingroup$
$frac 1 log x$ is not generally the same as $log frac 1 x $
$endgroup$
– J. W. Tanner
55 mins ago
$begingroup$
Did you mean $x=mathbf 2^-2/3$ ?
$endgroup$
– J. W. Tanner
40 mins ago
$begingroup$
Did you mean $x=mathbf 2^-2/3$ ?
$endgroup$
– J. W. Tanner
40 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.
answered 56 mins ago
Eevee TrainerEevee Trainer
7,72621338
$endgroup$
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?
answered 50 mins ago
J. W. TannerJ. W. Tanner
3,0701320
$endgroup$
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
44 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
30 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.
answered 56 mins ago
Eevee TrainerEevee Trainer
7,72621338
$endgroup$
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.
answered 56 mins ago
Eevee TrainerEevee Trainer
7,72621338
$endgroup$
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.
answered 56 mins ago
Eevee TrainerEevee Trainer
7,72621338
$endgroup$
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 56 mins ago
Eevee TrainerEevee Trainer
7,72621338
answered 56 mins ago
Eevee TrainerEevee Trainer
7,72621338
answered 56 mins ago
Eevee TrainerEevee Trainer
7,72621338
answered 56 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?
answered 50 mins ago
J. W. TannerJ. W. Tanner
3,0701320
$endgroup$
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
44 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
30 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?
answered 50 mins ago
J. W. TannerJ. W. Tanner
3,0701320
$endgroup$
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
44 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
30 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?
answered 50 mins ago
J. W. TannerJ. W. Tanner
3,0701320
$endgroup$
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 50 mins ago
J. W. TannerJ. W. Tanner
3,0701320
answered 50 mins ago
J. W. TannerJ. W. Tanner
3,0701320
answered 50 mins ago
J. W. TannerJ. W. Tanner
3,0701320
answered 50 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
44 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
30 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
44 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
30 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
44 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
44 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
30 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
30 mins ago
add a comment |
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-algebra-precalculus, logarithms
$begingroup$
$frac 1 log x$ is not generally the same as $log frac 1 x $
$endgroup$
– J. W. Tanner
55 mins ago
$begingroup$
Did you mean $x=mathbf 2^-2/3$ ?
$endgroup$
– J. W. Tanner
40 mins ago