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Trick to remember the difference between $sqrt[b]a^b=a$ and $log_aa^b=b$?
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I always forget which operation on $a^b$ gives $a$ and which operation gives $b$.
The answer of course is that:
$sqrt[b]a^b=a$ and $log_aa^b=b$
Somehow I always confuse the two.
Question: how can I easily remember the difference between the two? Any mnemonic rules?
algebra-precalculus
asked Jul 22 at 20:05
GambitSquared
1,2801034
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up vote
0
down vote
favorite
I always forget which operation on $a^b$ gives $a$ and which operation gives $b$.
The answer of course is that:
$sqrt[b]a^b=a$ and $log_aa^b=b$
Somehow I always confuse the two.
Question: how can I easily remember the difference between the two? Any mnemonic rules?
algebra-precalculus
asked Jul 22 at 20:05
GambitSquared
1,2801034
The inverse of $x^n=y$ is $x=sqrt [n] y$. That because the "opposite" of raising something to a specific power is to take a specific root. The inverse of $b^x$ is take a specific base and raising it to a variable power. The inverse of that is to take the base. If you remember what the functions are and that they are inverses of basic ideas.... you shouldn't have problems.
– fleablood
Jul 22 at 20:37
@GambitSquared Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 10 at 23:25
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I always forget which operation on $a^b$ gives $a$ and which operation gives $b$.
The answer of course is that:
$sqrt[b]a^b=a$ and $log_aa^b=b$
Somehow I always confuse the two.
Question: how can I easily remember the difference between the two? Any mnemonic rules?
algebra-precalculus
asked Jul 22 at 20:05
GambitSquared
1,2801034
I always forget which operation on $a^b$ gives $a$ and which operation gives $b$.
The answer of course is that:
$sqrt[b]a^b=a$ and $log_aa^b=b$
Somehow I always confuse the two.
Question: how can I easily remember the difference between the two? Any mnemonic rules?
algebra-precalculus
asked Jul 22 at 20:05
GambitSquared
1,2801034
edited Jul 22 at 20:11
asked Jul 22 at 20:05
GambitSquared
1,2801034
asked Jul 22 at 20:05
GambitSquared
1,2801034
asked Jul 22 at 20:05
GambitSquared
1,2801034
1,2801034
The inverse of $x^n=y$ is $x=sqrt [n] y$. That because the "opposite" of raising something to a specific power is to take a specific root. The inverse of $b^x$ is take a specific base and raising it to a variable power. The inverse of that is to take the base. If you remember what the functions are and that they are inverses of basic ideas.... you shouldn't have problems.
– fleablood
Jul 22 at 20:37
@GambitSquared Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 10 at 23:25
add a comment |Â
The inverse of $x^n=y$ is $x=sqrt [n] y$. That because the "opposite" of raising something to a specific power is to take a specific root. The inverse of $b^x$ is take a specific base and raising it to a variable power. The inverse of that is to take the base. If you remember what the functions are and that they are inverses of basic ideas.... you shouldn't have problems.
– fleablood
Jul 22 at 20:37
@GambitSquared Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 10 at 23:25
The inverse of $x^n=y$ is $x=sqrt [n] y$. That because the "opposite" of raising something to a specific power is to take a specific root. The inverse of $b^x$ is take a specific base and raising it to a variable power. The inverse of that is to take the base. If you remember what the functions are and that they are inverses of basic ideas.... you shouldn't have problems.
– fleablood
Jul 22 at 20:37
The inverse of $x^n=y$ is $x=sqrt [n] y$. That because the "opposite" of raising something to a specific power is to take a specific root. The inverse of $b^x$ is take a specific base and raising it to a variable power. The inverse of that is to take the base. If you remember what the functions are and that they are inverses of basic ideas.... you shouldn't have problems.
– fleablood
Jul 22 at 20:37
@GambitSquared Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 10 at 23:25
@GambitSquared Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 10 at 23:25
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Wanted to comment, but I don't have enough reputation...
Both results come straightforwardly from the definitions. If you have them clear in your mind, you shouldn't have this problem... Indeed, the square root is, in some domain (for example, in $mathbbR_geq 0$), the inverse function of the power. This means precisely that for any $ainmathbbR_geq 0$ you have $sqrt[n]a^n=a$.
As for the logarithm with base $a$, again with some care about the domain, it is the inverse function of the power with base $a$, so, again, you get precisely your result.
Also a numerical example could do the trick. Just think about $sqrt[2]a^2$ (if you want, also give a numerical value to $a$, but choose it different from $b=2$; for examples, for $a=3$ you get the square root of 9, which is $a$). As for the logarithm, just think about the logarithm with base 10, which "counts" the number of zeroes of the powers of 10: $Log_10(10^b)=b$, because $10^b$ has $b$ zeroes.
answered Jul 22 at 20:23
dpskrz
565
It's a very nice answer, don't need to put that as a comment!
– gimusi
Jul 22 at 20:32
add a comment |Â
up vote
3
down vote
As a memnonic rule, recall that we usually write
$$sqrt[n]x=x^frac1n$$
therefore
$$sqrt[b]a^b=(a^b)^frac1b=a^frac b b =a^1 =a$$
For the $log$ just keep in mind the definition
$$log_a x=b iff a^b=x$$
answered Jul 22 at 20:07
gimusi
65.2k73583
2
Add to that that $log(a^b)=blog(a)$, and then, if our logarithm happens to be base $a$, we get $blog_a(a)=bcdot 1$.
– Arthur
Jul 22 at 20:10
@Arthur Nice observation!
– gimusi
Jul 22 at 20:11
2
Be carefully with the algebra $$log(x^2)ne 2log(x)$$ in the general case.
– Dr. Sonnhard Graubner
Jul 22 at 20:21
@Dr.SonnhardGraubner Yes of course we are taking as usual $a>0$ and $aneq 1$ but of course we need to be aware for the other particular case.
– gimusi
Jul 22 at 20:29
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Wanted to comment, but I don't have enough reputation...
Both results come straightforwardly from the definitions. If you have them clear in your mind, you shouldn't have this problem... Indeed, the square root is, in some domain (for example, in $mathbbR_geq 0$), the inverse function of the power. This means precisely that for any $ainmathbbR_geq 0$ you have $sqrt[n]a^n=a$.
As for the logarithm with base $a$, again with some care about the domain, it is the inverse function of the power with base $a$, so, again, you get precisely your result.
Also a numerical example could do the trick. Just think about $sqrt[2]a^2$ (if you want, also give a numerical value to $a$, but choose it different from $b=2$; for examples, for $a=3$ you get the square root of 9, which is $a$). As for the logarithm, just think about the logarithm with base 10, which "counts" the number of zeroes of the powers of 10: $Log_10(10^b)=b$, because $10^b$ has $b$ zeroes.
answered Jul 22 at 20:23
dpskrz
565
It's a very nice answer, don't need to put that as a comment!
– gimusi
Jul 22 at 20:32
add a comment |Â
up vote
3
down vote
accepted
Wanted to comment, but I don't have enough reputation...
Both results come straightforwardly from the definitions. If you have them clear in your mind, you shouldn't have this problem... Indeed, the square root is, in some domain (for example, in $mathbbR_geq 0$), the inverse function of the power. This means precisely that for any $ainmathbbR_geq 0$ you have $sqrt[n]a^n=a$.
As for the logarithm with base $a$, again with some care about the domain, it is the inverse function of the power with base $a$, so, again, you get precisely your result.
Also a numerical example could do the trick. Just think about $sqrt[2]a^2$ (if you want, also give a numerical value to $a$, but choose it different from $b=2$; for examples, for $a=3$ you get the square root of 9, which is $a$). As for the logarithm, just think about the logarithm with base 10, which "counts" the number of zeroes of the powers of 10: $Log_10(10^b)=b$, because $10^b$ has $b$ zeroes.
answered Jul 22 at 20:23
dpskrz
565
It's a very nice answer, don't need to put that as a comment!
– gimusi
Jul 22 at 20:32
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Wanted to comment, but I don't have enough reputation...
Both results come straightforwardly from the definitions. If you have them clear in your mind, you shouldn't have this problem... Indeed, the square root is, in some domain (for example, in $mathbbR_geq 0$), the inverse function of the power. This means precisely that for any $ainmathbbR_geq 0$ you have $sqrt[n]a^n=a$.
As for the logarithm with base $a$, again with some care about the domain, it is the inverse function of the power with base $a$, so, again, you get precisely your result.
Also a numerical example could do the trick. Just think about $sqrt[2]a^2$ (if you want, also give a numerical value to $a$, but choose it different from $b=2$; for examples, for $a=3$ you get the square root of 9, which is $a$). As for the logarithm, just think about the logarithm with base 10, which "counts" the number of zeroes of the powers of 10: $Log_10(10^b)=b$, because $10^b$ has $b$ zeroes.
answered Jul 22 at 20:23
dpskrz
565
Wanted to comment, but I don't have enough reputation...
Both results come straightforwardly from the definitions. If you have them clear in your mind, you shouldn't have this problem... Indeed, the square root is, in some domain (for example, in $mathbbR_geq 0$), the inverse function of the power. This means precisely that for any $ainmathbbR_geq 0$ you have $sqrt[n]a^n=a$.
As for the logarithm with base $a$, again with some care about the domain, it is the inverse function of the power with base $a$, so, again, you get precisely your result.
Also a numerical example could do the trick. Just think about $sqrt[2]a^2$ (if you want, also give a numerical value to $a$, but choose it different from $b=2$; for examples, for $a=3$ you get the square root of 9, which is $a$). As for the logarithm, just think about the logarithm with base 10, which "counts" the number of zeroes of the powers of 10: $Log_10(10^b)=b$, because $10^b$ has $b$ zeroes.
answered Jul 22 at 20:23
dpskrz
565
edited Jul 22 at 21:28
answered Jul 22 at 20:23
dpskrz
565
answered Jul 22 at 20:23
dpskrz
565
answered Jul 22 at 20:23
dpskrz
565
565
It's a very nice answer, don't need to put that as a comment!
– gimusi
Jul 22 at 20:32
add a comment |Â
It's a very nice answer, don't need to put that as a comment!
– gimusi
Jul 22 at 20:32
It's a very nice answer, don't need to put that as a comment!
– gimusi
Jul 22 at 20:32
It's a very nice answer, don't need to put that as a comment!
– gimusi
Jul 22 at 20:32
add a comment |Â
up vote
3
down vote
As a memnonic rule, recall that we usually write
$$sqrt[n]x=x^frac1n$$
therefore
$$sqrt[b]a^b=(a^b)^frac1b=a^frac b b =a^1 =a$$
For the $log$ just keep in mind the definition
$$log_a x=b iff a^b=x$$
answered Jul 22 at 20:07
gimusi
65.2k73583
2
Add to that that $log(a^b)=blog(a)$, and then, if our logarithm happens to be base $a$, we get $blog_a(a)=bcdot 1$.
– Arthur
Jul 22 at 20:10
@Arthur Nice observation!
– gimusi
Jul 22 at 20:11
2
Be carefully with the algebra $$log(x^2)ne 2log(x)$$ in the general case.
– Dr. Sonnhard Graubner
Jul 22 at 20:21
@Dr.SonnhardGraubner Yes of course we are taking as usual $a>0$ and $aneq 1$ but of course we need to be aware for the other particular case.
– gimusi
Jul 22 at 20:29
add a comment |Â
up vote
3
down vote
As a memnonic rule, recall that we usually write
$$sqrt[n]x=x^frac1n$$
therefore
$$sqrt[b]a^b=(a^b)^frac1b=a^frac b b =a^1 =a$$
For the $log$ just keep in mind the definition
$$log_a x=b iff a^b=x$$
answered Jul 22 at 20:07
gimusi
65.2k73583
2
Add to that that $log(a^b)=blog(a)$, and then, if our logarithm happens to be base $a$, we get $blog_a(a)=bcdot 1$.
– Arthur
Jul 22 at 20:10
@Arthur Nice observation!
– gimusi
Jul 22 at 20:11
2
Be carefully with the algebra $$log(x^2)ne 2log(x)$$ in the general case.
– Dr. Sonnhard Graubner
Jul 22 at 20:21
@Dr.SonnhardGraubner Yes of course we are taking as usual $a>0$ and $aneq 1$ but of course we need to be aware for the other particular case.
– gimusi
Jul 22 at 20:29
add a comment |Â
up vote
3
down vote
up vote
3
down vote
As a memnonic rule, recall that we usually write
$$sqrt[n]x=x^frac1n$$
therefore
$$sqrt[b]a^b=(a^b)^frac1b=a^frac b b =a^1 =a$$
For the $log$ just keep in mind the definition
$$log_a x=b iff a^b=x$$
answered Jul 22 at 20:07
gimusi
65.2k73583
As a memnonic rule, recall that we usually write
$$sqrt[n]x=x^frac1n$$
therefore
$$sqrt[b]a^b=(a^b)^frac1b=a^frac b b =a^1 =a$$
For the $log$ just keep in mind the definition
$$log_a x=b iff a^b=x$$
answered Jul 22 at 20:07
gimusi
65.2k73583
answered Jul 22 at 20:07
gimusi
65.2k73583
answered Jul 22 at 20:07
gimusi
65.2k73583
answered Jul 22 at 20:07
gimusi
65.2k73583
65.2k73583
2
Add to that that $log(a^b)=blog(a)$, and then, if our logarithm happens to be base $a$, we get $blog_a(a)=bcdot 1$.
– Arthur
Jul 22 at 20:10
@Arthur Nice observation!
– gimusi
Jul 22 at 20:11
2
Be carefully with the algebra $$log(x^2)ne 2log(x)$$ in the general case.
– Dr. Sonnhard Graubner
Jul 22 at 20:21
@Dr.SonnhardGraubner Yes of course we are taking as usual $a>0$ and $aneq 1$ but of course we need to be aware for the other particular case.
– gimusi
Jul 22 at 20:29
add a comment |Â
2
Add to that that $log(a^b)=blog(a)$, and then, if our logarithm happens to be base $a$, we get $blog_a(a)=bcdot 1$.
– Arthur
Jul 22 at 20:10
@Arthur Nice observation!
– gimusi
Jul 22 at 20:11
2
Be carefully with the algebra $$log(x^2)ne 2log(x)$$ in the general case.
– Dr. Sonnhard Graubner
Jul 22 at 20:21
@Dr.SonnhardGraubner Yes of course we are taking as usual $a>0$ and $aneq 1$ but of course we need to be aware for the other particular case.
– gimusi
Jul 22 at 20:29
2
2
Add to that that $log(a^b)=blog(a)$, and then, if our logarithm happens to be base $a$, we get $blog_a(a)=bcdot 1$.
– Arthur
Jul 22 at 20:10
Add to that that $log(a^b)=blog(a)$, and then, if our logarithm happens to be base $a$, we get $blog_a(a)=bcdot 1$.
– Arthur
Jul 22 at 20:10
@Arthur Nice observation!
– gimusi
Jul 22 at 20:11
@Arthur Nice observation!
– gimusi
Jul 22 at 20:11
2
2
Be carefully with the algebra $$log(x^2)ne 2log(x)$$ in the general case.
– Dr. Sonnhard Graubner
Jul 22 at 20:21
Be carefully with the algebra $$log(x^2)ne 2log(x)$$ in the general case.
– Dr. Sonnhard Graubner
Jul 22 at 20:21
@Dr.SonnhardGraubner Yes of course we are taking as usual $a>0$ and $aneq 1$ but of course we need to be aware for the other particular case.
– gimusi
Jul 22 at 20:29
@Dr.SonnhardGraubner Yes of course we are taking as usual $a>0$ and $aneq 1$ but of course we need to be aware for the other particular case.
– gimusi
Jul 22 at 20:29
add a comment |Â
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The inverse of $x^n=y$ is $x=sqrt [n] y$. That because the "opposite" of raising something to a specific power is to take a specific root. The inverse of $b^x$ is take a specific base and raising it to a variable power. The inverse of that is to take the base. If you remember what the functions are and that they are inverses of basic ideas.... you shouldn't have problems.
– fleablood
Jul 22 at 20:37
@GambitSquared Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 10 at 23:25