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Solve: $2^xBigl(2^x-1Bigl) + 2^x-1Bigl(2^x-1 -1 Bigl) + … + 2^x-99Bigl(2^x-99 - 1Bigl) = 0$
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The question says to find the value of $x$ if, $$2^xBigl(2^x-1Bigl) + 2^x-1Bigl(2^x-1 -1 Bigl) + .... + 2^x-99Bigl(2^x-99 - 1 Bigl)= 0$$
My approach:
I rewrote the expression as,
$$2^xBigl(2^x-1Bigl) + frac2^x2Bigl(frac2^x2 -1 Bigl) + .... + frac2^x2^99 Bigl(frac2^x2^99 - 1 Bigl)= 0$$
I then took $bigl(2^xbigl)$ common and wrote it as,
$$2^x Biggl[ Bigl(2^x - 1Bigl) + frac12^1Bigl(2^x -2^1Bigl) + frac12^2Bigl(2^x - 2^2Bigl) + ;ldots + frac12^99 Bigl(2^x - 2^99Bigl)Biggl] = 0$$
After further simplification I got,
$$frac2^x2^99 Biggl[ Bigl(2^xcdot2^99 - 2^99Bigl) + Bigl(2^x cdot 2^98 - 2^99Bigl) + ldots + bigl(2^x -2^99bigl)Biggl] = 0$$
Taking $-2^99$ common I got,
$$-2^x Biggl[ Bigl( 2^x+99 + 2^x+98 + ldots + 2^x+2 + 2^x+1 + 2^x Bigl)Biggl]= 0$$
Now the inside can be expressed as $$sum ^ n= 99 _n=1 a_n$$ Where $a_n$ are the terms of the GP.
Thus we can see that either $$-2^x= 0$$ Or, $$sum ^ n= 99 _n=1 a_n = 0$$
Since the first condition is not poossible, thus, $$sum ^ n= 99 _n=1 a_n = 0$$
So,
$$2^x + 99 Biggl(frac1-frac12^1001-frac12 Biggl) = 0$$
Either way once I solve this, I am not getting an answer that is even in the options. The answers are all in the form of logarithmic expressions.
Any help would be appreciated. We have to find the value of $x$.
sequences-and-series algebra-precalculus logarithms
edited Jul 28 at 19:11
greedoid
26.1k93473
asked Jul 28 at 12:15
Prakhar Nagpal
428315
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up vote
5
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favorite
The question says to find the value of $x$ if, $$2^xBigl(2^x-1Bigl) + 2^x-1Bigl(2^x-1 -1 Bigl) + .... + 2^x-99Bigl(2^x-99 - 1 Bigl)= 0$$
My approach:
I rewrote the expression as,
$$2^xBigl(2^x-1Bigl) + frac2^x2Bigl(frac2^x2 -1 Bigl) + .... + frac2^x2^99 Bigl(frac2^x2^99 - 1 Bigl)= 0$$
I then took $bigl(2^xbigl)$ common and wrote it as,
$$2^x Biggl[ Bigl(2^x - 1Bigl) + frac12^1Bigl(2^x -2^1Bigl) + frac12^2Bigl(2^x - 2^2Bigl) + ;ldots + frac12^99 Bigl(2^x - 2^99Bigl)Biggl] = 0$$
After further simplification I got,
$$frac2^x2^99 Biggl[ Bigl(2^xcdot2^99 - 2^99Bigl) + Bigl(2^x cdot 2^98 - 2^99Bigl) + ldots + bigl(2^x -2^99bigl)Biggl] = 0$$
Taking $-2^99$ common I got,
$$-2^x Biggl[ Bigl( 2^x+99 + 2^x+98 + ldots + 2^x+2 + 2^x+1 + 2^x Bigl)Biggl]= 0$$
Now the inside can be expressed as $$sum ^ n= 99 _n=1 a_n$$ Where $a_n$ are the terms of the GP.
Thus we can see that either $$-2^x= 0$$ Or, $$sum ^ n= 99 _n=1 a_n = 0$$
Since the first condition is not poossible, thus, $$sum ^ n= 99 _n=1 a_n = 0$$
So,
$$2^x + 99 Biggl(frac1-frac12^1001-frac12 Biggl) = 0$$
Either way once I solve this, I am not getting an answer that is even in the options. The answers are all in the form of logarithmic expressions.
Any help would be appreciated. We have to find the value of $x$.
sequences-and-series algebra-precalculus logarithms
edited Jul 28 at 19:11
greedoid
26.1k93473
asked Jul 28 at 12:15
Prakhar Nagpal
428315
I understand now, how to reach at the answer, but can someone also explain what I have done wrong?
– Prakhar Nagpal
Jul 28 at 12:31
"can someone also explain what I have done wrong?" At the step "Taking $−2^99$ common I got", you forgot all the negative terms.
– Did
Jul 28 at 13:19
add a comment |Â
up vote
5
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favorite
up vote
5
down vote
favorite
The question says to find the value of $x$ if, $$2^xBigl(2^x-1Bigl) + 2^x-1Bigl(2^x-1 -1 Bigl) + .... + 2^x-99Bigl(2^x-99 - 1 Bigl)= 0$$
My approach:
I rewrote the expression as,
$$2^xBigl(2^x-1Bigl) + frac2^x2Bigl(frac2^x2 -1 Bigl) + .... + frac2^x2^99 Bigl(frac2^x2^99 - 1 Bigl)= 0$$
I then took $bigl(2^xbigl)$ common and wrote it as,
$$2^x Biggl[ Bigl(2^x - 1Bigl) + frac12^1Bigl(2^x -2^1Bigl) + frac12^2Bigl(2^x - 2^2Bigl) + ;ldots + frac12^99 Bigl(2^x - 2^99Bigl)Biggl] = 0$$
After further simplification I got,
$$frac2^x2^99 Biggl[ Bigl(2^xcdot2^99 - 2^99Bigl) + Bigl(2^x cdot 2^98 - 2^99Bigl) + ldots + bigl(2^x -2^99bigl)Biggl] = 0$$
Taking $-2^99$ common I got,
$$-2^x Biggl[ Bigl( 2^x+99 + 2^x+98 + ldots + 2^x+2 + 2^x+1 + 2^x Bigl)Biggl]= 0$$
Now the inside can be expressed as $$sum ^ n= 99 _n=1 a_n$$ Where $a_n$ are the terms of the GP.
Thus we can see that either $$-2^x= 0$$ Or, $$sum ^ n= 99 _n=1 a_n = 0$$
Since the first condition is not poossible, thus, $$sum ^ n= 99 _n=1 a_n = 0$$
So,
$$2^x + 99 Biggl(frac1-frac12^1001-frac12 Biggl) = 0$$
Either way once I solve this, I am not getting an answer that is even in the options. The answers are all in the form of logarithmic expressions.
Any help would be appreciated. We have to find the value of $x$.
sequences-and-series algebra-precalculus logarithms
edited Jul 28 at 19:11
greedoid
26.1k93473
asked Jul 28 at 12:15
Prakhar Nagpal
428315
The question says to find the value of $x$ if, $$2^xBigl(2^x-1Bigl) + 2^x-1Bigl(2^x-1 -1 Bigl) + .... + 2^x-99Bigl(2^x-99 - 1 Bigl)= 0$$
My approach:
I rewrote the expression as,
$$2^xBigl(2^x-1Bigl) + frac2^x2Bigl(frac2^x2 -1 Bigl) + .... + frac2^x2^99 Bigl(frac2^x2^99 - 1 Bigl)= 0$$
I then took $bigl(2^xbigl)$ common and wrote it as,
$$2^x Biggl[ Bigl(2^x - 1Bigl) + frac12^1Bigl(2^x -2^1Bigl) + frac12^2Bigl(2^x - 2^2Bigl) + ;ldots + frac12^99 Bigl(2^x - 2^99Bigl)Biggl] = 0$$
After further simplification I got,
$$frac2^x2^99 Biggl[ Bigl(2^xcdot2^99 - 2^99Bigl) + Bigl(2^x cdot 2^98 - 2^99Bigl) + ldots + bigl(2^x -2^99bigl)Biggl] = 0$$
Taking $-2^99$ common I got,
$$-2^x Biggl[ Bigl( 2^x+99 + 2^x+98 + ldots + 2^x+2 + 2^x+1 + 2^x Bigl)Biggl]= 0$$
Now the inside can be expressed as $$sum ^ n= 99 _n=1 a_n$$ Where $a_n$ are the terms of the GP.
Thus we can see that either $$-2^x= 0$$ Or, $$sum ^ n= 99 _n=1 a_n = 0$$
Since the first condition is not poossible, thus, $$sum ^ n= 99 _n=1 a_n = 0$$
So,
$$2^x + 99 Biggl(frac1-frac12^1001-frac12 Biggl) = 0$$
Either way once I solve this, I am not getting an answer that is even in the options. The answers are all in the form of logarithmic expressions.
Any help would be appreciated. We have to find the value of $x$.
sequences-and-series algebra-precalculus logarithms
edited Jul 28 at 19:11
greedoid
26.1k93473
asked Jul 28 at 12:15
Prakhar Nagpal
428315
edited Jul 28 at 19:11
greedoid
26.1k93473
edited Jul 28 at 19:11
greedoid
26.1k93473
edited Jul 28 at 19:11
greedoid
26.1k93473
26.1k93473
asked Jul 28 at 12:15
Prakhar Nagpal
428315
asked Jul 28 at 12:15
Prakhar Nagpal
428315
asked Jul 28 at 12:15
Prakhar Nagpal
428315
428315
I understand now, how to reach at the answer, but can someone also explain what I have done wrong?
– Prakhar Nagpal
Jul 28 at 12:31
"can someone also explain what I have done wrong?" At the step "Taking $−2^99$ common I got", you forgot all the negative terms.
– Did
Jul 28 at 13:19
add a comment |Â
I understand now, how to reach at the answer, but can someone also explain what I have done wrong?
– Prakhar Nagpal
Jul 28 at 12:31
"can someone also explain what I have done wrong?" At the step "Taking $−2^99$ common I got", you forgot all the negative terms.
– Did
Jul 28 at 13:19
I understand now, how to reach at the answer, but can someone also explain what I have done wrong?
– Prakhar Nagpal
Jul 28 at 12:31
I understand now, how to reach at the answer, but can someone also explain what I have done wrong?
– Prakhar Nagpal
Jul 28 at 12:31
"can someone also explain what I have done wrong?" At the step "Taking $−2^99$ common I got", you forgot all the negative terms.
– Did
Jul 28 at 13:19
"can someone also explain what I have done wrong?" At the step "Taking $−2^99$ common I got", you forgot all the negative terms.
– Did
Jul 28 at 13:19
add a comment |Â
5 Answers
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accepted
$$2^xBigl(2^x-1Bigl) + 2^x-1Bigl(2^x-1 -1 Bigl) + .... + 2^x-99Bigl(2^x-99 - 1 Bigl)= $$
$$2^2x-2^x + 2^2x-2-2^x-1 + .... + 2^2x-198-2^x-99= $$
$$(2^2x+ 2^2x-2 .... + 2^2x-198)-(2^x +2^x-1 + .... + 2^x-99)= $$
$$2^2x-198(2^198+ 2^196 .... + 2^0)-2^x-99(2^99 +2^98 + .... + 2^0)= $$
$$2^2x-1982^200-1over 2^2-1 -2^x-992^100-1over 2-1=0 $$
If we cancel this with $2^x-99(2^100-1)$ we get
$$2^x-992^100+1over 3 -1=0 $$
So $$2^x-99 = 3over 2^100+1$$ and thus $$x= 99+log_2 3over 2^100+1$$
answered Jul 28 at 12:28
greedoid
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beginalign
2^xBigl(2^x-1Bigl) + 2^x-1Bigl(2^x-1 -1 Bigl) + .... + 2^x-99Bigl(2^x-99 - 1 Bigl) &= 0 \
sum_i=0^992^x-ileft(2^x-i-1right) &= 0 \
sum_i=0^99left[left(2^x-iright)^2-2^x-iright] &= 0 \
sum_i=0^99left(2^x-iright)^2-sum_i=0^992^x-i &= 0 \
sum_i=0^992^2x-2i&=sum_i=0^992^x-i \
dfrac2^2xsum_i=0^992^2i&=dfrac2^xsum_i=0^992^i \
2^x&=dfracsum_i=0^992^2isum_i=0^992^i \
&= dfrac4^100-14-1cdotdfrac2-12^100-1 \
&= dfrac4^100-13left(2^100-1right)\
x &= log_2left(dfrac4^100-13left(2^100-1right)right)
endalign
answered Jul 28 at 12:29
Karn Watcharasupat
3,7992426
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From scratch, the quantity you are looking at is $$sum_k=0^99 2^2(x-k)-2^x-k=4^xsum_k=0^994^-k-2^xsum_k=0^992^-k=4^xcdotfrac4^-100-1-frac34-2^xfrac2^-100-1-frac12=\=2^xleft(2^xcdot frac4-4^-993-2+2^-99right)$$
That quantity is $0$ if and ony if $$2^x=frac3cdot 24cdotfrac1-2^-1001-4^-100=frac32(1+2^-100)\ x=fracln 3-ln(1+2^-100)ln 2-1$$
answered Jul 28 at 12:27
Saucy O'Path
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Here, you forgot red:
$$2^x Biggl[ Bigl(2^x - 1Bigl) + frac12^1cdot colorred2Bigl(2^x -2^1Bigl) + frac12^2colorred2^2Bigl(2^x - 2^2Bigl) + ;ldots + frac12^99colorred2^99 Bigl(2^x - 2^99Bigl)Biggl] = 0$$
answered Jul 28 at 12:39
greedoid
26.1k93473
But I have done the same thing in my answer, i simply wrote it as $Bigl (2^x -1bigl) + frac 12^1Bigl (frac2^x2 - 1Bigl).... $. I simplified that expression in a later step.
– Prakhar Nagpal
Jul 28 at 12:48
This is what I see and I think you didn't take it in consideration in later steps.
– greedoid
Jul 28 at 13:52
Alright. I will check my solution again.
– Prakhar Nagpal
Jul 28 at 14:33
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Hint.
$$
2^2xsum_k=0^992^-2k =2^xsum_k=0^992^-k
$$
hence
$$
2^x = fracsum_k=0^992^-ksum_k=0^992^-2k = 1.5to x = 0.5849625007211562
$$
answered Jul 28 at 12:27
Cesareo
5,6412412
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$$2^xBigl(2^x-1Bigl) + 2^x-1Bigl(2^x-1 -1 Bigl) + .... + 2^x-99Bigl(2^x-99 - 1 Bigl)= $$
$$2^2x-2^x + 2^2x-2-2^x-1 + .... + 2^2x-198-2^x-99= $$
$$(2^2x+ 2^2x-2 .... + 2^2x-198)-(2^x +2^x-1 + .... + 2^x-99)= $$
$$2^2x-198(2^198+ 2^196 .... + 2^0)-2^x-99(2^99 +2^98 + .... + 2^0)= $$
$$2^2x-1982^200-1over 2^2-1 -2^x-992^100-1over 2-1=0 $$
If we cancel this with $2^x-99(2^100-1)$ we get
$$2^x-992^100+1over 3 -1=0 $$
So $$2^x-99 = 3over 2^100+1$$ and thus $$x= 99+log_2 3over 2^100+1$$
answered Jul 28 at 12:28
greedoid
26.1k93473
add a comment |Â
up vote
1
down vote
accepted
$$2^xBigl(2^x-1Bigl) + 2^x-1Bigl(2^x-1 -1 Bigl) + .... + 2^x-99Bigl(2^x-99 - 1 Bigl)= $$
$$2^2x-2^x + 2^2x-2-2^x-1 + .... + 2^2x-198-2^x-99= $$
$$(2^2x+ 2^2x-2 .... + 2^2x-198)-(2^x +2^x-1 + .... + 2^x-99)= $$
$$2^2x-198(2^198+ 2^196 .... + 2^0)-2^x-99(2^99 +2^98 + .... + 2^0)= $$
$$2^2x-1982^200-1over 2^2-1 -2^x-992^100-1over 2-1=0 $$
If we cancel this with $2^x-99(2^100-1)$ we get
$$2^x-992^100+1over 3 -1=0 $$
So $$2^x-99 = 3over 2^100+1$$ and thus $$x= 99+log_2 3over 2^100+1$$
answered Jul 28 at 12:28
greedoid
26.1k93473
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$$2^xBigl(2^x-1Bigl) + 2^x-1Bigl(2^x-1 -1 Bigl) + .... + 2^x-99Bigl(2^x-99 - 1 Bigl)= $$
$$2^2x-2^x + 2^2x-2-2^x-1 + .... + 2^2x-198-2^x-99= $$
$$(2^2x+ 2^2x-2 .... + 2^2x-198)-(2^x +2^x-1 + .... + 2^x-99)= $$
$$2^2x-198(2^198+ 2^196 .... + 2^0)-2^x-99(2^99 +2^98 + .... + 2^0)= $$
$$2^2x-1982^200-1over 2^2-1 -2^x-992^100-1over 2-1=0 $$
If we cancel this with $2^x-99(2^100-1)$ we get
$$2^x-992^100+1over 3 -1=0 $$
So $$2^x-99 = 3over 2^100+1$$ and thus $$x= 99+log_2 3over 2^100+1$$
answered Jul 28 at 12:28
greedoid
26.1k93473
$$2^xBigl(2^x-1Bigl) + 2^x-1Bigl(2^x-1 -1 Bigl) + .... + 2^x-99Bigl(2^x-99 - 1 Bigl)= $$
$$2^2x-2^x + 2^2x-2-2^x-1 + .... + 2^2x-198-2^x-99= $$
$$(2^2x+ 2^2x-2 .... + 2^2x-198)-(2^x +2^x-1 + .... + 2^x-99)= $$
$$2^2x-198(2^198+ 2^196 .... + 2^0)-2^x-99(2^99 +2^98 + .... + 2^0)= $$
$$2^2x-1982^200-1over 2^2-1 -2^x-992^100-1over 2-1=0 $$
If we cancel this with $2^x-99(2^100-1)$ we get
$$2^x-992^100+1over 3 -1=0 $$
So $$2^x-99 = 3over 2^100+1$$ and thus $$x= 99+log_2 3over 2^100+1$$
answered Jul 28 at 12:28
greedoid
26.1k93473
edited Jul 28 at 12:40
answered Jul 28 at 12:28
greedoid
26.1k93473
answered Jul 28 at 12:28
greedoid
26.1k93473
answered Jul 28 at 12:28
greedoid
26.1k93473
26.1k93473
add a comment |Â
add a comment |Â
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beginalign
2^xBigl(2^x-1Bigl) + 2^x-1Bigl(2^x-1 -1 Bigl) + .... + 2^x-99Bigl(2^x-99 - 1 Bigl) &= 0 \
sum_i=0^992^x-ileft(2^x-i-1right) &= 0 \
sum_i=0^99left[left(2^x-iright)^2-2^x-iright] &= 0 \
sum_i=0^99left(2^x-iright)^2-sum_i=0^992^x-i &= 0 \
sum_i=0^992^2x-2i&=sum_i=0^992^x-i \
dfrac2^2xsum_i=0^992^2i&=dfrac2^xsum_i=0^992^i \
2^x&=dfracsum_i=0^992^2isum_i=0^992^i \
&= dfrac4^100-14-1cdotdfrac2-12^100-1 \
&= dfrac4^100-13left(2^100-1right)\
x &= log_2left(dfrac4^100-13left(2^100-1right)right)
endalign
answered Jul 28 at 12:29
Karn Watcharasupat
3,7992426
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up vote
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beginalign
2^xBigl(2^x-1Bigl) + 2^x-1Bigl(2^x-1 -1 Bigl) + .... + 2^x-99Bigl(2^x-99 - 1 Bigl) &= 0 \
sum_i=0^992^x-ileft(2^x-i-1right) &= 0 \
sum_i=0^99left[left(2^x-iright)^2-2^x-iright] &= 0 \
sum_i=0^99left(2^x-iright)^2-sum_i=0^992^x-i &= 0 \
sum_i=0^992^2x-2i&=sum_i=0^992^x-i \
dfrac2^2xsum_i=0^992^2i&=dfrac2^xsum_i=0^992^i \
2^x&=dfracsum_i=0^992^2isum_i=0^992^i \
&= dfrac4^100-14-1cdotdfrac2-12^100-1 \
&= dfrac4^100-13left(2^100-1right)\
x &= log_2left(dfrac4^100-13left(2^100-1right)right)
endalign
answered Jul 28 at 12:29
Karn Watcharasupat
3,7992426
add a comment |Â
up vote
0
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up vote
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beginalign
2^xBigl(2^x-1Bigl) + 2^x-1Bigl(2^x-1 -1 Bigl) + .... + 2^x-99Bigl(2^x-99 - 1 Bigl) &= 0 \
sum_i=0^992^x-ileft(2^x-i-1right) &= 0 \
sum_i=0^99left[left(2^x-iright)^2-2^x-iright] &= 0 \
sum_i=0^99left(2^x-iright)^2-sum_i=0^992^x-i &= 0 \
sum_i=0^992^2x-2i&=sum_i=0^992^x-i \
dfrac2^2xsum_i=0^992^2i&=dfrac2^xsum_i=0^992^i \
2^x&=dfracsum_i=0^992^2isum_i=0^992^i \
&= dfrac4^100-14-1cdotdfrac2-12^100-1 \
&= dfrac4^100-13left(2^100-1right)\
x &= log_2left(dfrac4^100-13left(2^100-1right)right)
endalign
answered Jul 28 at 12:29
Karn Watcharasupat
3,7992426
beginalign
2^xBigl(2^x-1Bigl) + 2^x-1Bigl(2^x-1 -1 Bigl) + .... + 2^x-99Bigl(2^x-99 - 1 Bigl) &= 0 \
sum_i=0^992^x-ileft(2^x-i-1right) &= 0 \
sum_i=0^99left[left(2^x-iright)^2-2^x-iright] &= 0 \
sum_i=0^99left(2^x-iright)^2-sum_i=0^992^x-i &= 0 \
sum_i=0^992^2x-2i&=sum_i=0^992^x-i \
dfrac2^2xsum_i=0^992^2i&=dfrac2^xsum_i=0^992^i \
2^x&=dfracsum_i=0^992^2isum_i=0^992^i \
&= dfrac4^100-14-1cdotdfrac2-12^100-1 \
&= dfrac4^100-13left(2^100-1right)\
x &= log_2left(dfrac4^100-13left(2^100-1right)right)
endalign
answered Jul 28 at 12:29
Karn Watcharasupat
3,7992426
answered Jul 28 at 12:29
Karn Watcharasupat
3,7992426
answered Jul 28 at 12:29
Karn Watcharasupat
3,7992426
answered Jul 28 at 12:29
Karn Watcharasupat
3,7992426
3,7992426
add a comment |Â
add a comment |Â
up vote
0
down vote
From scratch, the quantity you are looking at is $$sum_k=0^99 2^2(x-k)-2^x-k=4^xsum_k=0^994^-k-2^xsum_k=0^992^-k=4^xcdotfrac4^-100-1-frac34-2^xfrac2^-100-1-frac12=\=2^xleft(2^xcdot frac4-4^-993-2+2^-99right)$$
That quantity is $0$ if and ony if $$2^x=frac3cdot 24cdotfrac1-2^-1001-4^-100=frac32(1+2^-100)\ x=fracln 3-ln(1+2^-100)ln 2-1$$
answered Jul 28 at 12:27
Saucy O'Path
2,394217
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up vote
0
down vote
From scratch, the quantity you are looking at is $$sum_k=0^99 2^2(x-k)-2^x-k=4^xsum_k=0^994^-k-2^xsum_k=0^992^-k=4^xcdotfrac4^-100-1-frac34-2^xfrac2^-100-1-frac12=\=2^xleft(2^xcdot frac4-4^-993-2+2^-99right)$$
That quantity is $0$ if and ony if $$2^x=frac3cdot 24cdotfrac1-2^-1001-4^-100=frac32(1+2^-100)\ x=fracln 3-ln(1+2^-100)ln 2-1$$
answered Jul 28 at 12:27
Saucy O'Path
2,394217
add a comment |Â
up vote
0
down vote
up vote
0
down vote
From scratch, the quantity you are looking at is $$sum_k=0^99 2^2(x-k)-2^x-k=4^xsum_k=0^994^-k-2^xsum_k=0^992^-k=4^xcdotfrac4^-100-1-frac34-2^xfrac2^-100-1-frac12=\=2^xleft(2^xcdot frac4-4^-993-2+2^-99right)$$
That quantity is $0$ if and ony if $$2^x=frac3cdot 24cdotfrac1-2^-1001-4^-100=frac32(1+2^-100)\ x=fracln 3-ln(1+2^-100)ln 2-1$$
answered Jul 28 at 12:27
Saucy O'Path
2,394217
From scratch, the quantity you are looking at is $$sum_k=0^99 2^2(x-k)-2^x-k=4^xsum_k=0^994^-k-2^xsum_k=0^992^-k=4^xcdotfrac4^-100-1-frac34-2^xfrac2^-100-1-frac12=\=2^xleft(2^xcdot frac4-4^-993-2+2^-99right)$$
That quantity is $0$ if and ony if $$2^x=frac3cdot 24cdotfrac1-2^-1001-4^-100=frac32(1+2^-100)\ x=fracln 3-ln(1+2^-100)ln 2-1$$
answered Jul 28 at 12:27
Saucy O'Path
2,394217
edited Jul 28 at 12:33
answered Jul 28 at 12:27
Saucy O'Path
2,394217
answered Jul 28 at 12:27
Saucy O'Path
2,394217
answered Jul 28 at 12:27
Saucy O'Path
2,394217
2,394217
add a comment |Â
add a comment |Â
up vote
0
down vote
Here, you forgot red:
$$2^x Biggl[ Bigl(2^x - 1Bigl) + frac12^1cdot colorred2Bigl(2^x -2^1Bigl) + frac12^2colorred2^2Bigl(2^x - 2^2Bigl) + ;ldots + frac12^99colorred2^99 Bigl(2^x - 2^99Bigl)Biggl] = 0$$
answered Jul 28 at 12:39
greedoid
26.1k93473
But I have done the same thing in my answer, i simply wrote it as $Bigl (2^x -1bigl) + frac 12^1Bigl (frac2^x2 - 1Bigl).... $. I simplified that expression in a later step.
– Prakhar Nagpal
Jul 28 at 12:48
This is what I see and I think you didn't take it in consideration in later steps.
– greedoid
Jul 28 at 13:52
Alright. I will check my solution again.
– Prakhar Nagpal
Jul 28 at 14:33
add a comment |Â
up vote
0
down vote
Here, you forgot red:
$$2^x Biggl[ Bigl(2^x - 1Bigl) + frac12^1cdot colorred2Bigl(2^x -2^1Bigl) + frac12^2colorred2^2Bigl(2^x - 2^2Bigl) + ;ldots + frac12^99colorred2^99 Bigl(2^x - 2^99Bigl)Biggl] = 0$$
answered Jul 28 at 12:39
greedoid
26.1k93473
But I have done the same thing in my answer, i simply wrote it as $Bigl (2^x -1bigl) + frac 12^1Bigl (frac2^x2 - 1Bigl).... $. I simplified that expression in a later step.
– Prakhar Nagpal
Jul 28 at 12:48
This is what I see and I think you didn't take it in consideration in later steps.
– greedoid
Jul 28 at 13:52
Alright. I will check my solution again.
– Prakhar Nagpal
Jul 28 at 14:33
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here, you forgot red:
$$2^x Biggl[ Bigl(2^x - 1Bigl) + frac12^1cdot colorred2Bigl(2^x -2^1Bigl) + frac12^2colorred2^2Bigl(2^x - 2^2Bigl) + ;ldots + frac12^99colorred2^99 Bigl(2^x - 2^99Bigl)Biggl] = 0$$
answered Jul 28 at 12:39
greedoid
26.1k93473
Here, you forgot red:
$$2^x Biggl[ Bigl(2^x - 1Bigl) + frac12^1cdot colorred2Bigl(2^x -2^1Bigl) + frac12^2colorred2^2Bigl(2^x - 2^2Bigl) + ;ldots + frac12^99colorred2^99 Bigl(2^x - 2^99Bigl)Biggl] = 0$$
answered Jul 28 at 12:39
greedoid
26.1k93473
answered Jul 28 at 12:39
greedoid
26.1k93473
answered Jul 28 at 12:39
greedoid
26.1k93473
answered Jul 28 at 12:39
greedoid
26.1k93473
26.1k93473
But I have done the same thing in my answer, i simply wrote it as $Bigl (2^x -1bigl) + frac 12^1Bigl (frac2^x2 - 1Bigl).... $. I simplified that expression in a later step.
– Prakhar Nagpal
Jul 28 at 12:48
This is what I see and I think you didn't take it in consideration in later steps.
– greedoid
Jul 28 at 13:52
Alright. I will check my solution again.
– Prakhar Nagpal
Jul 28 at 14:33
add a comment |Â
But I have done the same thing in my answer, i simply wrote it as $Bigl (2^x -1bigl) + frac 12^1Bigl (frac2^x2 - 1Bigl).... $. I simplified that expression in a later step.
– Prakhar Nagpal
Jul 28 at 12:48
This is what I see and I think you didn't take it in consideration in later steps.
– greedoid
Jul 28 at 13:52
Alright. I will check my solution again.
– Prakhar Nagpal
Jul 28 at 14:33
But I have done the same thing in my answer, i simply wrote it as $Bigl (2^x -1bigl) + frac 12^1Bigl (frac2^x2 - 1Bigl).... $. I simplified that expression in a later step.
– Prakhar Nagpal
Jul 28 at 12:48
But I have done the same thing in my answer, i simply wrote it as $Bigl (2^x -1bigl) + frac 12^1Bigl (frac2^x2 - 1Bigl).... $. I simplified that expression in a later step.
– Prakhar Nagpal
Jul 28 at 12:48
This is what I see and I think you didn't take it in consideration in later steps.
– greedoid
Jul 28 at 13:52
This is what I see and I think you didn't take it in consideration in later steps.
– greedoid
Jul 28 at 13:52
Alright. I will check my solution again.
– Prakhar Nagpal
Jul 28 at 14:33
Alright. I will check my solution again.
– Prakhar Nagpal
Jul 28 at 14:33
add a comment |Â
up vote
0
down vote
Hint.
$$
2^2xsum_k=0^992^-2k =2^xsum_k=0^992^-k
$$
hence
$$
2^x = fracsum_k=0^992^-ksum_k=0^992^-2k = 1.5to x = 0.5849625007211562
$$
answered Jul 28 at 12:27
Cesareo
5,6412412
add a comment |Â
up vote
0
down vote
Hint.
$$
2^2xsum_k=0^992^-2k =2^xsum_k=0^992^-k
$$
hence
$$
2^x = fracsum_k=0^992^-ksum_k=0^992^-2k = 1.5to x = 0.5849625007211562
$$
answered Jul 28 at 12:27
Cesareo
5,6412412
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint.
$$
2^2xsum_k=0^992^-2k =2^xsum_k=0^992^-k
$$
hence
$$
2^x = fracsum_k=0^992^-ksum_k=0^992^-2k = 1.5to x = 0.5849625007211562
$$
answered Jul 28 at 12:27
Cesareo
5,6412412
Hint.
$$
2^2xsum_k=0^992^-2k =2^xsum_k=0^992^-k
$$
hence
$$
2^x = fracsum_k=0^992^-ksum_k=0^992^-2k = 1.5to x = 0.5849625007211562
$$
answered Jul 28 at 12:27
Cesareo
5,6412412
edited Jul 28 at 12:42
answered Jul 28 at 12:27
Cesareo
5,6412412
answered Jul 28 at 12:27
Cesareo
5,6412412
answered Jul 28 at 12:27
Cesareo
5,6412412
5,6412412
add a comment |Â
add a comment |Â
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I understand now, how to reach at the answer, but can someone also explain what I have done wrong?
– Prakhar Nagpal
Jul 28 at 12:31
"can someone also explain what I have done wrong?" At the step "Taking $−2^99$ common I got", you forgot all the negative terms.
– Did
Jul 28 at 13:19