X^K+2 split up exponent rules
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I am working on a problem and I am confused if exponents can be split up in the manner below.
$$x^{k+2} = x^{k+1}+x^{k+1} = 2x^{k+1} $$
I apologize if this is a simple question I tried to look up exponent rules but I couldn't find exponents + a number.
algebra-precalculus discrete-mathematics
asked Dec 3 '18 at 18:08
Arthur GreenArthur Green
796
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add a comment |
$begingroup$
I am working on a problem and I am confused if exponents can be split up in the manner below.
$$x^{k+2} = x^{k+1}+x^{k+1} = 2x^{k+1} $$
I apologize if this is a simple question I tried to look up exponent rules but I couldn't find exponents + a number.
algebra-precalculus discrete-mathematics
asked Dec 3 '18 at 18:08
Arthur GreenArthur Green
796
$endgroup$
$begingroup$
Arthur.$x^{2k+1}=x^{2k}x^1= xx^{2k}$
$endgroup$
– Peter Szilas
Dec 3 '18 at 18:14
add a comment |
$begingroup$
I am working on a problem and I am confused if exponents can be split up in the manner below.
$$x^{k+2} = x^{k+1}+x^{k+1} = 2x^{k+1} $$
I apologize if this is a simple question I tried to look up exponent rules but I couldn't find exponents + a number.
algebra-precalculus discrete-mathematics
asked Dec 3 '18 at 18:08
Arthur GreenArthur Green
796
$endgroup$
I am working on a problem and I am confused if exponents can be split up in the manner below.
$$x^{k+2} = x^{k+1}+x^{k+1} = 2x^{k+1} $$
I apologize if this is a simple question I tried to look up exponent rules but I couldn't find exponents + a number.
algebra-precalculus discrete-mathematics
algebra-precalculus discrete-mathematics
asked Dec 3 '18 at 18:08
Arthur GreenArthur Green
796
asked Dec 3 '18 at 18:08
Arthur GreenArthur Green
796
asked Dec 3 '18 at 18:08
Arthur GreenArthur Green
796
asked Dec 3 '18 at 18:08
Arthur GreenArthur Green
796
asked Dec 3 '18 at 18:08
Arthur GreenArthur Green
796
796
$begingroup$
Arthur.$x^{2k+1}=x^{2k}x^1= xx^{2k}$
$endgroup$
– Peter Szilas
Dec 3 '18 at 18:14
add a comment |
$begingroup$
Arthur.$x^{2k+1}=x^{2k}x^1= xx^{2k}$
$endgroup$
– Peter Szilas
Dec 3 '18 at 18:14
$begingroup$
Arthur.$x^{2k+1}=x^{2k}x^1= xx^{2k}$
$endgroup$
– Peter Szilas
Dec 3 '18 at 18:14
$begingroup$
Arthur.$x^{2k+1}=x^{2k}x^1= xx^{2k}$
$endgroup$
– Peter Szilas
Dec 3 '18 at 18:14
add a comment |
2 Answers
2
active
oldest
votes
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No, they’re not broken up like that. You need to use
$$a^bcdot a^c = a^{b+c}$$
and vice-versa. In general, you can try to create a simple expression to see if your way is correct. For example, notice
$$2^{2+2} = 16 color{red}{neq 2^2+2^2 = 8}$$
answered Dec 3 '18 at 18:11
KM101KM101
6,0251525
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add a comment |
$begingroup$
No, they definitely cannot be split up like that. To my knowledge there isn't any general law for $x^a + x^b = text{something}$.
A few handy exponent laws can be found at http://mathworld.wolfram.com/ExponentLaws.html
answered Dec 3 '18 at 18:11
Eevee TrainerEevee Trainer
6,61311237
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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active
oldest
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$begingroup$
No, they’re not broken up like that. You need to use
$$a^bcdot a^c = a^{b+c}$$
and vice-versa. In general, you can try to create a simple expression to see if your way is correct. For example, notice
$$2^{2+2} = 16 color{red}{neq 2^2+2^2 = 8}$$
answered Dec 3 '18 at 18:11
KM101KM101
6,0251525
$endgroup$
add a comment |
$begingroup$
No, they’re not broken up like that. You need to use
$$a^bcdot a^c = a^{b+c}$$
and vice-versa. In general, you can try to create a simple expression to see if your way is correct. For example, notice
$$2^{2+2} = 16 color{red}{neq 2^2+2^2 = 8}$$
answered Dec 3 '18 at 18:11
KM101KM101
6,0251525
$endgroup$
add a comment |
$begingroup$
No, they’re not broken up like that. You need to use
$$a^bcdot a^c = a^{b+c}$$
and vice-versa. In general, you can try to create a simple expression to see if your way is correct. For example, notice
$$2^{2+2} = 16 color{red}{neq 2^2+2^2 = 8}$$
answered Dec 3 '18 at 18:11
KM101KM101
6,0251525
$endgroup$
No, they’re not broken up like that. You need to use
$$a^bcdot a^c = a^{b+c}$$
and vice-versa. In general, you can try to create a simple expression to see if your way is correct. For example, notice
$$2^{2+2} = 16 color{red}{neq 2^2+2^2 = 8}$$
answered Dec 3 '18 at 18:11
KM101KM101
6,0251525
answered Dec 3 '18 at 18:11
KM101KM101
6,0251525
answered Dec 3 '18 at 18:11
KM101KM101
6,0251525
answered Dec 3 '18 at 18:11
KM101KM101
6,0251525
6,0251525
add a comment |
add a comment |
$begingroup$
No, they definitely cannot be split up like that. To my knowledge there isn't any general law for $x^a + x^b = text{something}$.
A few handy exponent laws can be found at http://mathworld.wolfram.com/ExponentLaws.html
answered Dec 3 '18 at 18:11
Eevee TrainerEevee Trainer
6,61311237
$endgroup$
add a comment |
$begingroup$
No, they definitely cannot be split up like that. To my knowledge there isn't any general law for $x^a + x^b = text{something}$.
A few handy exponent laws can be found at http://mathworld.wolfram.com/ExponentLaws.html
answered Dec 3 '18 at 18:11
Eevee TrainerEevee Trainer
6,61311237
$endgroup$
add a comment |
$begingroup$
No, they definitely cannot be split up like that. To my knowledge there isn't any general law for $x^a + x^b = text{something}$.
A few handy exponent laws can be found at http://mathworld.wolfram.com/ExponentLaws.html
answered Dec 3 '18 at 18:11
Eevee TrainerEevee Trainer
6,61311237
$endgroup$
No, they definitely cannot be split up like that. To my knowledge there isn't any general law for $x^a + x^b = text{something}$.
A few handy exponent laws can be found at http://mathworld.wolfram.com/ExponentLaws.html
answered Dec 3 '18 at 18:11
Eevee TrainerEevee Trainer
6,61311237
answered Dec 3 '18 at 18:11
Eevee TrainerEevee Trainer
6,61311237
answered Dec 3 '18 at 18:11
Eevee TrainerEevee Trainer
6,61311237
answered Dec 3 '18 at 18:11
Eevee TrainerEevee Trainer
6,61311237
6,61311237
add a comment |
add a comment |
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$begingroup$
Arthur.$x^{2k+1}=x^{2k}x^1= xx^{2k}$
$endgroup$
– Peter Szilas
Dec 3 '18 at 18:14