Mixing
Simplification with exponents
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I'm currently revising an exam about channel coding in telecommunications and we have a question where we need to isolate a variable $u$ in terms of another variable $q$. Currently, I am stuck with a rather difficult expression to simplify or at least I can't see what the trick is (if there is one). Here is the said expression:
$$log_2big(q^q(1-q)^{1-q}big) = log_2bigg((frac{qu}{1-qu})^qbigg)$$
Being perfectly honest, I know that the $log_2$ simplifies on each side but having $u$ in both the numerator and denominator of the fraction which has an overall exponent, I'm quite out of my comfort zone mathematically.
Thanks in advance!
calculus exponentiation coding-theory
asked Jan 24 at 19:42
Max MichelMax Michel
32
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add a comment |
$begingroup$
I'm currently revising an exam about channel coding in telecommunications and we have a question where we need to isolate a variable $u$ in terms of another variable $q$. Currently, I am stuck with a rather difficult expression to simplify or at least I can't see what the trick is (if there is one). Here is the said expression:
$$log_2big(q^q(1-q)^{1-q}big) = log_2bigg((frac{qu}{1-qu})^qbigg)$$
Being perfectly honest, I know that the $log_2$ simplifies on each side but having $u$ in both the numerator and denominator of the fraction which has an overall exponent, I'm quite out of my comfort zone mathematically.
Thanks in advance!
calculus exponentiation coding-theory
asked Jan 24 at 19:42
Max MichelMax Michel
32
$endgroup$
$begingroup$
What is $u$? It doesn't appear in the left hand side
$endgroup$
– pwerth
Jan 24 at 19:47
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That was one of my problems ^^
$endgroup$
– Max Michel
Jan 24 at 19:58
add a comment |
$begingroup$
I'm currently revising an exam about channel coding in telecommunications and we have a question where we need to isolate a variable $u$ in terms of another variable $q$. Currently, I am stuck with a rather difficult expression to simplify or at least I can't see what the trick is (if there is one). Here is the said expression:
$$log_2big(q^q(1-q)^{1-q}big) = log_2bigg((frac{qu}{1-qu})^qbigg)$$
Being perfectly honest, I know that the $log_2$ simplifies on each side but having $u$ in both the numerator and denominator of the fraction which has an overall exponent, I'm quite out of my comfort zone mathematically.
Thanks in advance!
calculus exponentiation coding-theory
asked Jan 24 at 19:42
Max MichelMax Michel
32
$endgroup$
I'm currently revising an exam about channel coding in telecommunications and we have a question where we need to isolate a variable $u$ in terms of another variable $q$. Currently, I am stuck with a rather difficult expression to simplify or at least I can't see what the trick is (if there is one). Here is the said expression:
$$log_2big(q^q(1-q)^{1-q}big) = log_2bigg((frac{qu}{1-qu})^qbigg)$$
Being perfectly honest, I know that the $log_2$ simplifies on each side but having $u$ in both the numerator and denominator of the fraction which has an overall exponent, I'm quite out of my comfort zone mathematically.
Thanks in advance!
calculus exponentiation coding-theory
calculus exponentiation coding-theory
asked Jan 24 at 19:42
Max MichelMax Michel
32
asked Jan 24 at 19:42
Max MichelMax Michel
32
asked Jan 24 at 19:42
Max MichelMax Michel
32
asked Jan 24 at 19:42
Max MichelMax Michel
32
asked Jan 24 at 19:42
Max MichelMax Michel
32
32
$begingroup$
What is $u$? It doesn't appear in the left hand side
$endgroup$
– pwerth
Jan 24 at 19:47
$begingroup$
That was one of my problems ^^
$endgroup$
– Max Michel
Jan 24 at 19:58
add a comment |
$begingroup$
What is $u$? It doesn't appear in the left hand side
$endgroup$
– pwerth
Jan 24 at 19:47
$begingroup$
That was one of my problems ^^
$endgroup$
– Max Michel
Jan 24 at 19:58
$begingroup$
What is $u$? It doesn't appear in the left hand side
$endgroup$
– pwerth
Jan 24 at 19:47
$begingroup$
What is $u$? It doesn't appear in the left hand side
$endgroup$
– pwerth
Jan 24 at 19:47
$begingroup$
That was one of my problems ^^
$endgroup$
– Max Michel
Jan 24 at 19:58
$begingroup$
That was one of my problems ^^
$endgroup$
– Max Michel
Jan 24 at 19:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since we have on both sides the base $2$ we get
$$q^q(1-q)^{1-q}=left(frac {qu}{1-qu}right)^q$$ Let's say $$A=q^q(1-q)^{1-q}$$ then $$a^{1/q}=frac{qu}{1-qu}$$ so $$A^{1/q}-qA^{1/q}=qu$$ and $$A^{1/q}-quA^{1/q}=qu$$ so $$A^{1/q}=u(q+qA^{1/q})$$ and we obtain
$$u=frac{A^{1/q}}{q+qA^{1/q}}$$
answered Jan 24 at 19:52
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Since we have on both sides the base $2$ we get
$$q^q(1-q)^{1-q}=left(frac {qu}{1-qu}right)^q$$ Let's say $$A=q^q(1-q)^{1-q}$$ then $$a^{1/q}=frac{qu}{1-qu}$$ so $$A^{1/q}-qA^{1/q}=qu$$ and $$A^{1/q}-quA^{1/q}=qu$$ so $$A^{1/q}=u(q+qA^{1/q})$$ and we obtain
$$u=frac{A^{1/q}}{q+qA^{1/q}}$$
answered Jan 24 at 19:52
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
add a comment |
$begingroup$
Since we have on both sides the base $2$ we get
$$q^q(1-q)^{1-q}=left(frac {qu}{1-qu}right)^q$$ Let's say $$A=q^q(1-q)^{1-q}$$ then $$a^{1/q}=frac{qu}{1-qu}$$ so $$A^{1/q}-qA^{1/q}=qu$$ and $$A^{1/q}-quA^{1/q}=qu$$ so $$A^{1/q}=u(q+qA^{1/q})$$ and we obtain
$$u=frac{A^{1/q}}{q+qA^{1/q}}$$
answered Jan 24 at 19:52
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
add a comment |
$begingroup$
Since we have on both sides the base $2$ we get
$$q^q(1-q)^{1-q}=left(frac {qu}{1-qu}right)^q$$ Let's say $$A=q^q(1-q)^{1-q}$$ then $$a^{1/q}=frac{qu}{1-qu}$$ so $$A^{1/q}-qA^{1/q}=qu$$ and $$A^{1/q}-quA^{1/q}=qu$$ so $$A^{1/q}=u(q+qA^{1/q})$$ and we obtain
$$u=frac{A^{1/q}}{q+qA^{1/q}}$$
answered Jan 24 at 19:52
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
Since we have on both sides the base $2$ we get
$$q^q(1-q)^{1-q}=left(frac {qu}{1-qu}right)^q$$ Let's say $$A=q^q(1-q)^{1-q}$$ then $$a^{1/q}=frac{qu}{1-qu}$$ so $$A^{1/q}-qA^{1/q}=qu$$ and $$A^{1/q}-quA^{1/q}=qu$$ so $$A^{1/q}=u(q+qA^{1/q})$$ and we obtain
$$u=frac{A^{1/q}}{q+qA^{1/q}}$$
answered Jan 24 at 19:52
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
answered Jan 24 at 19:52
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
answered Jan 24 at 19:52
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
answered Jan 24 at 19:52
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
77.9k42866
add a comment |
add a comment |
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$begingroup$
What is $u$? It doesn't appear in the left hand side
$endgroup$
– pwerth
Jan 24 at 19:47
$begingroup$
That was one of my problems ^^
$endgroup$
– Max Michel
Jan 24 at 19:58