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Solve the logarithmic equation
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I have logarithmic problem:
$$fraclog :_10left(1:+:frac12+frac14+frac18+...right)log _10left(2+frac23+frac29+...right)cdot left(log _2left(3right)+log _3left(4right)+log _16left(3right)+...+log _2^2nleft(3right)+..right)$$
I solved first part:
$$fraclog :_10left(2right)log _10left(3right)cdot left(log _2left(3right)+log _3left(4right)+log _16left(3right)+...+log _2^2nleft(3right)+..right)$$
But can't understand second part.
Answers:
$A=2, B=-1, C=-2, D=frac12$
summation logarithms
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up vote
0
down vote
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I have logarithmic problem:
$$fraclog :_10left(1:+:frac12+frac14+frac18+...right)log _10left(2+frac23+frac29+...right)cdot left(log _2left(3right)+log _3left(4right)+log _16left(3right)+...+log _2^2nleft(3right)+..right)$$
I solved first part:
$$fraclog :_10left(2right)log _10left(3right)cdot left(log _2left(3right)+log _3left(4right)+log _16left(3right)+...+log _2^2nleft(3right)+..right)$$
But can't understand second part.
Answers:
$A=2, B=-1, C=-2, D=frac12$
summation logarithms
1
Did you mean $log_4 (3)$ in the second part?
– jiaminglimjm
Jul 24 at 4:33
@jiaminglimjm no i didn't mean that
– user578407
Jul 24 at 4:36
Don't understand what? And how can $log_3(4)$ be "correct"? Correct about what? Why is that any more correct than any other mathematical expression? Anyway what's your question?
– fleablood
Jul 24 at 4:54
1
Well, OP seems to mean that the second term in the expression is not a mistake, even though neither first or second seem to fit the remaining term pattern...
– gt6989b
Jul 24 at 4:57
"no i didn't mean that" Are you sure you don't mean that? Where did you get this problem from. Do you have any reason to think it wasn't a typo. It doesn't make any sense that it's not a typo.
– fleablood
Jul 24 at 4:59
 |Â
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have logarithmic problem:
$$fraclog :_10left(1:+:frac12+frac14+frac18+...right)log _10left(2+frac23+frac29+...right)cdot left(log _2left(3right)+log _3left(4right)+log _16left(3right)+...+log _2^2nleft(3right)+..right)$$
I solved first part:
$$fraclog :_10left(2right)log _10left(3right)cdot left(log _2left(3right)+log _3left(4right)+log _16left(3right)+...+log _2^2nleft(3right)+..right)$$
But can't understand second part.
Answers:
$A=2, B=-1, C=-2, D=frac12$
summation logarithms
I have logarithmic problem:
$$fraclog :_10left(1:+:frac12+frac14+frac18+...right)log _10left(2+frac23+frac29+...right)cdot left(log _2left(3right)+log _3left(4right)+log _16left(3right)+...+log _2^2nleft(3right)+..right)$$
I solved first part:
$$fraclog :_10left(2right)log _10left(3right)cdot left(log _2left(3right)+log _3left(4right)+log _16left(3right)+...+log _2^2nleft(3right)+..right)$$
But can't understand second part.
Answers:
$A=2, B=-1, C=-2, D=frac12$
summation logarithms
edited Jul 24 at 5:06
asked Jul 24 at 4:28
user578407
1
Did you mean $log_4 (3)$ in the second part?
– jiaminglimjm
Jul 24 at 4:33
@jiaminglimjm no i didn't mean that
– user578407
Jul 24 at 4:36
Don't understand what? And how can $log_3(4)$ be "correct"? Correct about what? Why is that any more correct than any other mathematical expression? Anyway what's your question?
– fleablood
Jul 24 at 4:54
1
Well, OP seems to mean that the second term in the expression is not a mistake, even though neither first or second seem to fit the remaining term pattern...
– gt6989b
Jul 24 at 4:57
"no i didn't mean that" Are you sure you don't mean that? Where did you get this problem from. Do you have any reason to think it wasn't a typo. It doesn't make any sense that it's not a typo.
– fleablood
Jul 24 at 4:59
 |Â
show 3 more comments
1
Did you mean $log_4 (3)$ in the second part?
– jiaminglimjm
Jul 24 at 4:33
@jiaminglimjm no i didn't mean that
– user578407
Jul 24 at 4:36
Don't understand what? And how can $log_3(4)$ be "correct"? Correct about what? Why is that any more correct than any other mathematical expression? Anyway what's your question?
– fleablood
Jul 24 at 4:54
1
Well, OP seems to mean that the second term in the expression is not a mistake, even though neither first or second seem to fit the remaining term pattern...
– gt6989b
Jul 24 at 4:57
"no i didn't mean that" Are you sure you don't mean that? Where did you get this problem from. Do you have any reason to think it wasn't a typo. It doesn't make any sense that it's not a typo.
– fleablood
Jul 24 at 4:59
1
1
Did you mean $log_4 (3)$ in the second part?
– jiaminglimjm
Jul 24 at 4:33
Did you mean $log_4 (3)$ in the second part?
– jiaminglimjm
Jul 24 at 4:33
@jiaminglimjm no i didn't mean that
– user578407
Jul 24 at 4:36
@jiaminglimjm no i didn't mean that
– user578407
Jul 24 at 4:36
Don't understand what? And how can $log_3(4)$ be "correct"? Correct about what? Why is that any more correct than any other mathematical expression? Anyway what's your question?
– fleablood
Jul 24 at 4:54
Don't understand what? And how can $log_3(4)$ be "correct"? Correct about what? Why is that any more correct than any other mathematical expression? Anyway what's your question?
– fleablood
Jul 24 at 4:54
1
1
Well, OP seems to mean that the second term in the expression is not a mistake, even though neither first or second seem to fit the remaining term pattern...
– gt6989b
Jul 24 at 4:57
Well, OP seems to mean that the second term in the expression is not a mistake, even though neither first or second seem to fit the remaining term pattern...
– gt6989b
Jul 24 at 4:57
"no i didn't mean that" Are you sure you don't mean that? Where did you get this problem from. Do you have any reason to think it wasn't a typo. It doesn't make any sense that it's not a typo.
– fleablood
Jul 24 at 4:59
"no i didn't mean that" Are you sure you don't mean that? Where did you get this problem from. Do you have any reason to think it wasn't a typo. It doesn't make any sense that it's not a typo.
– fleablood
Jul 24 at 4:59
 |Â
show 3 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
$$
S = sum_n=1^infty log_2^2n(3)
= sum_n=1^infty log_4^n(3)
= sum_n=1^infty fracln 3ln 4^n
= fracln 3ln 4 sum_n=1^infty frac1n
$$
which diverges.
answered Jul 24 at 4:56
gt6989b
30.3k22148
add a comment |Â
up vote
1
down vote
For the second part we can write for terms after the first two $$log_2^2n3=a_n\
left(2^2nright)^a_n=3\2^2na_n=3\2^a_n=3^frac 12n\
a_n=frac 12nlog_23$$
and all the terms after the first two become a diverging harmonic series.
Now that you have added the choices, it is clear the text expects the series to converge. Alpha agrees with me-you can see the screenshot below. I assume the book answer is A. All the terms are positive, so the result cannot be negative. The first two terms of the second part are greater than $1$, so the whole expression is greater than $2frac log_102log_103approx 1.26$ which rules out D. Even if it did converge, the irregularity of the first two terms would make me sure the result is not some nice number like $2$.
answered Jul 24 at 4:59
Ross Millikan
275k21186351
Thanx i appreciate it
– user578407
Jul 24 at 5:19
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$$
S = sum_n=1^infty log_2^2n(3)
= sum_n=1^infty log_4^n(3)
= sum_n=1^infty fracln 3ln 4^n
= fracln 3ln 4 sum_n=1^infty frac1n
$$
which diverges.
answered Jul 24 at 4:56
gt6989b
30.3k22148
add a comment |Â
up vote
2
down vote
$$
S = sum_n=1^infty log_2^2n(3)
= sum_n=1^infty log_4^n(3)
= sum_n=1^infty fracln 3ln 4^n
= fracln 3ln 4 sum_n=1^infty frac1n
$$
which diverges.
answered Jul 24 at 4:56
gt6989b
30.3k22148
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$$
S = sum_n=1^infty log_2^2n(3)
= sum_n=1^infty log_4^n(3)
= sum_n=1^infty fracln 3ln 4^n
= fracln 3ln 4 sum_n=1^infty frac1n
$$
which diverges.
answered Jul 24 at 4:56
gt6989b
30.3k22148
$$
S = sum_n=1^infty log_2^2n(3)
= sum_n=1^infty log_4^n(3)
= sum_n=1^infty fracln 3ln 4^n
= fracln 3ln 4 sum_n=1^infty frac1n
$$
which diverges.
answered Jul 24 at 4:56
gt6989b
30.3k22148
answered Jul 24 at 4:56
gt6989b
30.3k22148
answered Jul 24 at 4:56
gt6989b
30.3k22148
answered Jul 24 at 4:56
gt6989b
30.3k22148
30.3k22148
add a comment |Â
add a comment |Â
up vote
1
down vote
For the second part we can write for terms after the first two $$log_2^2n3=a_n\
left(2^2nright)^a_n=3\2^2na_n=3\2^a_n=3^frac 12n\
a_n=frac 12nlog_23$$
and all the terms after the first two become a diverging harmonic series.
Now that you have added the choices, it is clear the text expects the series to converge. Alpha agrees with me-you can see the screenshot below. I assume the book answer is A. All the terms are positive, so the result cannot be negative. The first two terms of the second part are greater than $1$, so the whole expression is greater than $2frac log_102log_103approx 1.26$ which rules out D. Even if it did converge, the irregularity of the first two terms would make me sure the result is not some nice number like $2$.
answered Jul 24 at 4:59
Ross Millikan
275k21186351
Thanx i appreciate it
– user578407
Jul 24 at 5:19
add a comment |Â
up vote
1
down vote
For the second part we can write for terms after the first two $$log_2^2n3=a_n\
left(2^2nright)^a_n=3\2^2na_n=3\2^a_n=3^frac 12n\
a_n=frac 12nlog_23$$
and all the terms after the first two become a diverging harmonic series.
Now that you have added the choices, it is clear the text expects the series to converge. Alpha agrees with me-you can see the screenshot below. I assume the book answer is A. All the terms are positive, so the result cannot be negative. The first two terms of the second part are greater than $1$, so the whole expression is greater than $2frac log_102log_103approx 1.26$ which rules out D. Even if it did converge, the irregularity of the first two terms would make me sure the result is not some nice number like $2$.
answered Jul 24 at 4:59
Ross Millikan
275k21186351
Thanx i appreciate it
– user578407
Jul 24 at 5:19
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For the second part we can write for terms after the first two $$log_2^2n3=a_n\
left(2^2nright)^a_n=3\2^2na_n=3\2^a_n=3^frac 12n\
a_n=frac 12nlog_23$$
and all the terms after the first two become a diverging harmonic series.
Now that you have added the choices, it is clear the text expects the series to converge. Alpha agrees with me-you can see the screenshot below. I assume the book answer is A. All the terms are positive, so the result cannot be negative. The first two terms of the second part are greater than $1$, so the whole expression is greater than $2frac log_102log_103approx 1.26$ which rules out D. Even if it did converge, the irregularity of the first two terms would make me sure the result is not some nice number like $2$.
answered Jul 24 at 4:59
Ross Millikan
275k21186351
For the second part we can write for terms after the first two $$log_2^2n3=a_n\
left(2^2nright)^a_n=3\2^2na_n=3\2^a_n=3^frac 12n\
a_n=frac 12nlog_23$$
and all the terms after the first two become a diverging harmonic series.
Now that you have added the choices, it is clear the text expects the series to converge. Alpha agrees with me-you can see the screenshot below. I assume the book answer is A. All the terms are positive, so the result cannot be negative. The first two terms of the second part are greater than $1$, so the whole expression is greater than $2frac log_102log_103approx 1.26$ which rules out D. Even if it did converge, the irregularity of the first two terms would make me sure the result is not some nice number like $2$.
answered Jul 24 at 4:59
Ross Millikan
275k21186351
edited Jul 24 at 5:19
answered Jul 24 at 4:59
Ross Millikan
275k21186351
answered Jul 24 at 4:59
Ross Millikan
275k21186351
answered Jul 24 at 4:59
Ross Millikan
275k21186351
275k21186351
Thanx i appreciate it
– user578407
Jul 24 at 5:19
add a comment |Â
Thanx i appreciate it
– user578407
Jul 24 at 5:19
Thanx i appreciate it
– user578407
Jul 24 at 5:19
Thanx i appreciate it
– user578407
Jul 24 at 5:19
add a comment |Â
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1
Did you mean $log_4 (3)$ in the second part?
– jiaminglimjm
Jul 24 at 4:33
@jiaminglimjm no i didn't mean that
– user578407
Jul 24 at 4:36
Don't understand what? And how can $log_3(4)$ be "correct"? Correct about what? Why is that any more correct than any other mathematical expression? Anyway what's your question?
– fleablood
Jul 24 at 4:54
1
Well, OP seems to mean that the second term in the expression is not a mistake, even though neither first or second seem to fit the remaining term pattern...
– gt6989b
Jul 24 at 4:57
"no i didn't mean that" Are you sure you don't mean that? Where did you get this problem from. Do you have any reason to think it wasn't a typo. It doesn't make any sense that it's not a typo.
– fleablood
Jul 24 at 4:59