Simplify $frac56logleft(frac54right) - frac16log(2)$ to $logleft(frac54right) - frac16logleft(frac52right)$
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I'm trying to bring this expression:
$$frac56logleft(frac54right) - frac16log(2)$$
To this one:
$$logleft(frac54right) - frac16logleft(frac52right)$$
Where $log$ is the natural algorithm.
I know the two expressions are equal (checked with wolfram) but I really can't find the correct passages... Could you please help me?
algebra-precalculus logarithms
edited Sep 22 at 10:53
user21820
36.7k441143
asked Sep 21 at 23:28
Alessio Martorana
1057
add a comment |Â
up vote
3
down vote
favorite
I'm trying to bring this expression:
$$frac56logleft(frac54right) - frac16log(2)$$
To this one:
$$logleft(frac54right) - frac16logleft(frac52right)$$
Where $log$ is the natural algorithm.
I know the two expressions are equal (checked with wolfram) but I really can't find the correct passages... Could you please help me?
algebra-precalculus logarithms
edited Sep 22 at 10:53
user21820
36.7k441143
asked Sep 21 at 23:28
Alessio Martorana
1057
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm trying to bring this expression:
$$frac56logleft(frac54right) - frac16log(2)$$
To this one:
$$logleft(frac54right) - frac16logleft(frac52right)$$
Where $log$ is the natural algorithm.
I know the two expressions are equal (checked with wolfram) but I really can't find the correct passages... Could you please help me?
algebra-precalculus logarithms
edited Sep 22 at 10:53
user21820
36.7k441143
asked Sep 21 at 23:28
Alessio Martorana
1057
I'm trying to bring this expression:
$$frac56logleft(frac54right) - frac16log(2)$$
To this one:
$$logleft(frac54right) - frac16logleft(frac52right)$$
Where $log$ is the natural algorithm.
I know the two expressions are equal (checked with wolfram) but I really can't find the correct passages... Could you please help me?
algebra-precalculus logarithms
algebra-precalculus logarithms
edited Sep 22 at 10:53
user21820
36.7k441143
asked Sep 21 at 23:28
Alessio Martorana
1057
edited Sep 22 at 10:53
user21820
36.7k441143
asked Sep 21 at 23:28
Alessio Martorana
1057
edited Sep 22 at 10:53
user21820
36.7k441143
edited Sep 22 at 10:53
user21820
36.7k441143
edited Sep 22 at 10:53
user21820
36.7k441143
36.7k441143
asked Sep 21 at 23:28
Alessio Martorana
1057
asked Sep 21 at 23:28
Alessio Martorana
1057
asked Sep 21 at 23:28
Alessio Martorana
1057
1057
add a comment |Â
add a comment |Â
5 Answers
5
active
oldest
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up vote
6
down vote
accepted
$frac56logleft(frac54right)-frac16log(2)=logleft(frac54right)-frac16logleft(frac54right)-frac16log(2)$
$= logleft(frac54right)-(frac16logleft(frac54right)+frac16log(2))$
$=logleft(frac54right)-frac16logleft(frac5cdot24right)$
$=logleft(frac54right)-frac16logleft(frac52right)$
I should add: Subtracting logs is equivalent to division. The reason for potential problems here is $fracfracabc = fracabcdot c$ and not $fraca(fracbc)$
answered Sep 22 at 0:15
Phil H
2,7412311
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up vote
6
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Hint: Begin with$$frac56logleft(frac54right)-frac16log(2)=logleft(frac54right)-frac16logleft(frac54right)-frac16log(2).$$
answered Sep 21 at 23:31
José Carlos Santos
126k17102189
Ty very much for the hint, I was on the right road, but missed the last passage that, imho, Phil stated very simply and clear
– Alessio Martorana
Sep 22 at 0:25
add a comment |Â
up vote
3
down vote
We have
$$frac56logleft(frac54right) - frac16log 2=overbracefrac56logleft(frac54right)+colorredfrac16logleft(frac54right) -overbracecolorredfrac16logleft(frac54right)- frac16log 2=$$
$$=logleft(frac54right)-frac16 logleft(frac52right)$$
indeed recall that $log A+ log B= log AB$ and therefore
$$-frac16logleft(frac54right)- frac16log 2=-frac16left[logleft(frac54right)+log 2right]=-frac16logleft(frac54cdot 2right)=-frac16logleft(frac52right)$$
answered Sep 22 at 0:16
gimusi
1
add a comment |Â
up vote
2
down vote
Alt. hint: Â by brute force, using only $,log fracab = log a - log b,$ and $,log a^n = n log a,$:
$$small
frac56logfrac54 - frac16log 2 = frac16left(5 log 5 - 5 log 4-log 2right) = frac16left(5 log 5 - 10 log 2-log 2right) = frac16left(5 log 5 - 11 log 2right)
$$
Now do the same for $,log frac54 - frac16log frac52,$ and compare.
answered Sep 22 at 0:19
dxiv
57.3k64799
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0
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$$fraccolorblue56log frac54 - frac16log 2 = fraccolorblue6-16logfrac54 - frac16log2 = log frac54 - frac16left(colorgreenlogfrac54 + log 2 right)$$$$ = log frac54 - frac16colorgreen log left( frac54 cdot 2 right) = log frac54 - frac16 log frac52 $$
answered Sep 22 at 0:09
trancelocation
6,2711516
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
$frac56logleft(frac54right)-frac16log(2)=logleft(frac54right)-frac16logleft(frac54right)-frac16log(2)$
$= logleft(frac54right)-(frac16logleft(frac54right)+frac16log(2))$
$=logleft(frac54right)-frac16logleft(frac5cdot24right)$
$=logleft(frac54right)-frac16logleft(frac52right)$
I should add: Subtracting logs is equivalent to division. The reason for potential problems here is $fracfracabc = fracabcdot c$ and not $fraca(fracbc)$
answered Sep 22 at 0:15
Phil H
2,7412311
add a comment |Â
up vote
6
down vote
accepted
$frac56logleft(frac54right)-frac16log(2)=logleft(frac54right)-frac16logleft(frac54right)-frac16log(2)$
$= logleft(frac54right)-(frac16logleft(frac54right)+frac16log(2))$
$=logleft(frac54right)-frac16logleft(frac5cdot24right)$
$=logleft(frac54right)-frac16logleft(frac52right)$
I should add: Subtracting logs is equivalent to division. The reason for potential problems here is $fracfracabc = fracabcdot c$ and not $fraca(fracbc)$
answered Sep 22 at 0:15
Phil H
2,7412311
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
$frac56logleft(frac54right)-frac16log(2)=logleft(frac54right)-frac16logleft(frac54right)-frac16log(2)$
$= logleft(frac54right)-(frac16logleft(frac54right)+frac16log(2))$
$=logleft(frac54right)-frac16logleft(frac5cdot24right)$
$=logleft(frac54right)-frac16logleft(frac52right)$
I should add: Subtracting logs is equivalent to division. The reason for potential problems here is $fracfracabc = fracabcdot c$ and not $fraca(fracbc)$
answered Sep 22 at 0:15
Phil H
2,7412311
$frac56logleft(frac54right)-frac16log(2)=logleft(frac54right)-frac16logleft(frac54right)-frac16log(2)$
$= logleft(frac54right)-(frac16logleft(frac54right)+frac16log(2))$
$=logleft(frac54right)-frac16logleft(frac5cdot24right)$
$=logleft(frac54right)-frac16logleft(frac52right)$
I should add: Subtracting logs is equivalent to division. The reason for potential problems here is $fracfracabc = fracabcdot c$ and not $fraca(fracbc)$
answered Sep 22 at 0:15
Phil H
2,7412311
edited Sep 22 at 1:45
answered Sep 22 at 0:15
Phil H
2,7412311
answered Sep 22 at 0:15
Phil H
2,7412311
answered Sep 22 at 0:15
Phil H
2,7412311
2,7412311
add a comment |Â
add a comment |Â
up vote
6
down vote
Hint: Begin with$$frac56logleft(frac54right)-frac16log(2)=logleft(frac54right)-frac16logleft(frac54right)-frac16log(2).$$
answered Sep 21 at 23:31
José Carlos Santos
126k17102189
Ty very much for the hint, I was on the right road, but missed the last passage that, imho, Phil stated very simply and clear
– Alessio Martorana
Sep 22 at 0:25
add a comment |Â
up vote
6
down vote
Hint: Begin with$$frac56logleft(frac54right)-frac16log(2)=logleft(frac54right)-frac16logleft(frac54right)-frac16log(2).$$
answered Sep 21 at 23:31
José Carlos Santos
126k17102189
Ty very much for the hint, I was on the right road, but missed the last passage that, imho, Phil stated very simply and clear
– Alessio Martorana
Sep 22 at 0:25
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Hint: Begin with$$frac56logleft(frac54right)-frac16log(2)=logleft(frac54right)-frac16logleft(frac54right)-frac16log(2).$$
answered Sep 21 at 23:31
José Carlos Santos
126k17102189
Hint: Begin with$$frac56logleft(frac54right)-frac16log(2)=logleft(frac54right)-frac16logleft(frac54right)-frac16log(2).$$
answered Sep 21 at 23:31
José Carlos Santos
126k17102189
answered Sep 21 at 23:31
José Carlos Santos
126k17102189
answered Sep 21 at 23:31
José Carlos Santos
126k17102189
answered Sep 21 at 23:31
José Carlos Santos
126k17102189
126k17102189
Ty very much for the hint, I was on the right road, but missed the last passage that, imho, Phil stated very simply and clear
– Alessio Martorana
Sep 22 at 0:25
add a comment |Â
Ty very much for the hint, I was on the right road, but missed the last passage that, imho, Phil stated very simply and clear
– Alessio Martorana
Sep 22 at 0:25
Ty very much for the hint, I was on the right road, but missed the last passage that, imho, Phil stated very simply and clear
– Alessio Martorana
Sep 22 at 0:25
Ty very much for the hint, I was on the right road, but missed the last passage that, imho, Phil stated very simply and clear
– Alessio Martorana
Sep 22 at 0:25
add a comment |Â
up vote
3
down vote
We have
$$frac56logleft(frac54right) - frac16log 2=overbracefrac56logleft(frac54right)+colorredfrac16logleft(frac54right) -overbracecolorredfrac16logleft(frac54right)- frac16log 2=$$
$$=logleft(frac54right)-frac16 logleft(frac52right)$$
indeed recall that $log A+ log B= log AB$ and therefore
$$-frac16logleft(frac54right)- frac16log 2=-frac16left[logleft(frac54right)+log 2right]=-frac16logleft(frac54cdot 2right)=-frac16logleft(frac52right)$$
answered Sep 22 at 0:16
gimusi
1
add a comment |Â
up vote
3
down vote
We have
$$frac56logleft(frac54right) - frac16log 2=overbracefrac56logleft(frac54right)+colorredfrac16logleft(frac54right) -overbracecolorredfrac16logleft(frac54right)- frac16log 2=$$
$$=logleft(frac54right)-frac16 logleft(frac52right)$$
indeed recall that $log A+ log B= log AB$ and therefore
$$-frac16logleft(frac54right)- frac16log 2=-frac16left[logleft(frac54right)+log 2right]=-frac16logleft(frac54cdot 2right)=-frac16logleft(frac52right)$$
answered Sep 22 at 0:16
gimusi
1
add a comment |Â
up vote
3
down vote
up vote
3
down vote
We have
$$frac56logleft(frac54right) - frac16log 2=overbracefrac56logleft(frac54right)+colorredfrac16logleft(frac54right) -overbracecolorredfrac16logleft(frac54right)- frac16log 2=$$
$$=logleft(frac54right)-frac16 logleft(frac52right)$$
indeed recall that $log A+ log B= log AB$ and therefore
$$-frac16logleft(frac54right)- frac16log 2=-frac16left[logleft(frac54right)+log 2right]=-frac16logleft(frac54cdot 2right)=-frac16logleft(frac52right)$$
answered Sep 22 at 0:16
gimusi
1
We have
$$frac56logleft(frac54right) - frac16log 2=overbracefrac56logleft(frac54right)+colorredfrac16logleft(frac54right) -overbracecolorredfrac16logleft(frac54right)- frac16log 2=$$
$$=logleft(frac54right)-frac16 logleft(frac52right)$$
indeed recall that $log A+ log B= log AB$ and therefore
$$-frac16logleft(frac54right)- frac16log 2=-frac16left[logleft(frac54right)+log 2right]=-frac16logleft(frac54cdot 2right)=-frac16logleft(frac52right)$$
answered Sep 22 at 0:16
gimusi
1
edited Sep 22 at 6:00
answered Sep 22 at 0:16
gimusi
1
answered Sep 22 at 0:16
gimusi
1
answered Sep 22 at 0:16
gimusi
1
1
add a comment |Â
add a comment |Â
up vote
2
down vote
Alt. hint: Â by brute force, using only $,log fracab = log a - log b,$ and $,log a^n = n log a,$:
$$small
frac56logfrac54 - frac16log 2 = frac16left(5 log 5 - 5 log 4-log 2right) = frac16left(5 log 5 - 10 log 2-log 2right) = frac16left(5 log 5 - 11 log 2right)
$$
Now do the same for $,log frac54 - frac16log frac52,$ and compare.
answered Sep 22 at 0:19
dxiv
57.3k64799
add a comment |Â
up vote
2
down vote
Alt. hint: Â by brute force, using only $,log fracab = log a - log b,$ and $,log a^n = n log a,$:
$$small
frac56logfrac54 - frac16log 2 = frac16left(5 log 5 - 5 log 4-log 2right) = frac16left(5 log 5 - 10 log 2-log 2right) = frac16left(5 log 5 - 11 log 2right)
$$
Now do the same for $,log frac54 - frac16log frac52,$ and compare.
answered Sep 22 at 0:19
dxiv
57.3k64799
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Alt. hint: Â by brute force, using only $,log fracab = log a - log b,$ and $,log a^n = n log a,$:
$$small
frac56logfrac54 - frac16log 2 = frac16left(5 log 5 - 5 log 4-log 2right) = frac16left(5 log 5 - 10 log 2-log 2right) = frac16left(5 log 5 - 11 log 2right)
$$
Now do the same for $,log frac54 - frac16log frac52,$ and compare.
answered Sep 22 at 0:19
dxiv
57.3k64799
Alt. hint: Â by brute force, using only $,log fracab = log a - log b,$ and $,log a^n = n log a,$:
$$small
frac56logfrac54 - frac16log 2 = frac16left(5 log 5 - 5 log 4-log 2right) = frac16left(5 log 5 - 10 log 2-log 2right) = frac16left(5 log 5 - 11 log 2right)
$$
Now do the same for $,log frac54 - frac16log frac52,$ and compare.
answered Sep 22 at 0:19
dxiv
57.3k64799
answered Sep 22 at 0:19
dxiv
57.3k64799
answered Sep 22 at 0:19
dxiv
57.3k64799
answered Sep 22 at 0:19
dxiv
57.3k64799
57.3k64799
add a comment |Â
add a comment |Â
up vote
0
down vote
$$fraccolorblue56log frac54 - frac16log 2 = fraccolorblue6-16logfrac54 - frac16log2 = log frac54 - frac16left(colorgreenlogfrac54 + log 2 right)$$$$ = log frac54 - frac16colorgreen log left( frac54 cdot 2 right) = log frac54 - frac16 log frac52 $$
answered Sep 22 at 0:09
trancelocation
6,2711516
add a comment |Â
up vote
0
down vote
$$fraccolorblue56log frac54 - frac16log 2 = fraccolorblue6-16logfrac54 - frac16log2 = log frac54 - frac16left(colorgreenlogfrac54 + log 2 right)$$$$ = log frac54 - frac16colorgreen log left( frac54 cdot 2 right) = log frac54 - frac16 log frac52 $$
answered Sep 22 at 0:09
trancelocation
6,2711516
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$fraccolorblue56log frac54 - frac16log 2 = fraccolorblue6-16logfrac54 - frac16log2 = log frac54 - frac16left(colorgreenlogfrac54 + log 2 right)$$$$ = log frac54 - frac16colorgreen log left( frac54 cdot 2 right) = log frac54 - frac16 log frac52 $$
answered Sep 22 at 0:09
trancelocation
6,2711516
$$fraccolorblue56log frac54 - frac16log 2 = fraccolorblue6-16logfrac54 - frac16log2 = log frac54 - frac16left(colorgreenlogfrac54 + log 2 right)$$$$ = log frac54 - frac16colorgreen log left( frac54 cdot 2 right) = log frac54 - frac16 log frac52 $$
answered Sep 22 at 0:09
trancelocation
6,2711516
answered Sep 22 at 0:09
trancelocation
6,2711516
answered Sep 22 at 0:09
trancelocation
6,2711516
answered Sep 22 at 0:09
trancelocation
6,2711516
6,2711516
add a comment |Â
add a comment |Â
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