Solve this : $ x^{frac 34 { log_2^2 x} + log_2 {(x-5/4) }} = sqrt 2 $
$begingroup$
The question : $$ x^{frac 34 { log_2^2 x} + log_2 {left(x-frac 54right) }} = sqrt 2 $$
My try at it : I think that the whole equation should be converted to log with base 2. What to do next?
logarithms
asked Apr 14 '17 at 14:33
Esha MukhopadhyayEsha Mukhopadhyay
315
$endgroup$
|
show 9 more comments
$begingroup$
The question : $$ x^{frac 34 { log_2^2 x} + log_2 {left(x-frac 54right) }} = sqrt 2 $$
My try at it : I think that the whole equation should be converted to log with base 2. What to do next?
logarithms
asked Apr 14 '17 at 14:33
Esha MukhopadhyayEsha Mukhopadhyay
315
$endgroup$
$begingroup$
Does $log^2 x$ mean $log(log x)$ or $(log x)^2$
$endgroup$
– DHMO
Apr 14 '17 at 14:39
1
$begingroup$
The title and the question have different equations in them. Which one do you mean?
$endgroup$
– AsafHaas
Apr 14 '17 at 14:40
$begingroup$
$ (log_2 x) ^ 2 $
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
@AsafHaas its the same. i hav copied it from the title.
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
By executing your idea and let $y=log_2 x$, we get $1/2=y(3y^2/4+log_2{(x-5/4)})$
$endgroup$
– didgogns
Apr 14 '17 at 14:43
|
show 9 more comments
$begingroup$
The question : $$ x^{frac 34 { log_2^2 x} + log_2 {left(x-frac 54right) }} = sqrt 2 $$
My try at it : I think that the whole equation should be converted to log with base 2. What to do next?
logarithms
asked Apr 14 '17 at 14:33
Esha MukhopadhyayEsha Mukhopadhyay
315
$endgroup$
The question : $$ x^{frac 34 { log_2^2 x} + log_2 {left(x-frac 54right) }} = sqrt 2 $$
My try at it : I think that the whole equation should be converted to log with base 2. What to do next?
logarithms
logarithms
asked Apr 14 '17 at 14:33
Esha MukhopadhyayEsha Mukhopadhyay
315
asked Apr 14 '17 at 14:33
Esha MukhopadhyayEsha Mukhopadhyay
315
edited Apr 14 '17 at 15:09
Esha Mukhopadhyay
asked Apr 14 '17 at 14:33
Esha MukhopadhyayEsha Mukhopadhyay
315
asked Apr 14 '17 at 14:33
Esha MukhopadhyayEsha Mukhopadhyay
315
asked Apr 14 '17 at 14:33
Esha MukhopadhyayEsha Mukhopadhyay
315
315
$begingroup$
Does $log^2 x$ mean $log(log x)$ or $(log x)^2$
$endgroup$
– DHMO
Apr 14 '17 at 14:39
1
$begingroup$
The title and the question have different equations in them. Which one do you mean?
$endgroup$
– AsafHaas
Apr 14 '17 at 14:40
$begingroup$
$ (log_2 x) ^ 2 $
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
@AsafHaas its the same. i hav copied it from the title.
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
By executing your idea and let $y=log_2 x$, we get $1/2=y(3y^2/4+log_2{(x-5/4)})$
$endgroup$
– didgogns
Apr 14 '17 at 14:43
|
show 9 more comments
$begingroup$
Does $log^2 x$ mean $log(log x)$ or $(log x)^2$
$endgroup$
– DHMO
Apr 14 '17 at 14:39
1
$begingroup$
The title and the question have different equations in them. Which one do you mean?
$endgroup$
– AsafHaas
Apr 14 '17 at 14:40
$begingroup$
$ (log_2 x) ^ 2 $
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
@AsafHaas its the same. i hav copied it from the title.
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
By executing your idea and let $y=log_2 x$, we get $1/2=y(3y^2/4+log_2{(x-5/4)})$
$endgroup$
– didgogns
Apr 14 '17 at 14:43
$begingroup$
Does $log^2 x$ mean $log(log x)$ or $(log x)^2$
$endgroup$
– DHMO
Apr 14 '17 at 14:39
$begingroup$
Does $log^2 x$ mean $log(log x)$ or $(log x)^2$
$endgroup$
– DHMO
Apr 14 '17 at 14:39
1
1
$begingroup$
The title and the question have different equations in them. Which one do you mean?
$endgroup$
– AsafHaas
Apr 14 '17 at 14:40
$begingroup$
The title and the question have different equations in them. Which one do you mean?
$endgroup$
– AsafHaas
Apr 14 '17 at 14:40
$begingroup$
$ (log_2 x) ^ 2 $
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
$ (log_2 x) ^ 2 $
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
@AsafHaas its the same. i hav copied it from the title.
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
@AsafHaas its the same. i hav copied it from the title.
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
By executing your idea and let $y=log_2 x$, we get $1/2=y(3y^2/4+log_2{(x-5/4)})$
$endgroup$
– didgogns
Apr 14 '17 at 14:43
$begingroup$
By executing your idea and let $y=log_2 x$, we get $1/2=y(3y^2/4+log_2{(x-5/4)})$
$endgroup$
– didgogns
Apr 14 '17 at 14:43
|
show 9 more comments
1 Answer
1
active
oldest
votes
$begingroup$
taking the logarithm on both sides of your equation we obatin
$$left(frac{3}{4}left(frac{ln(x)}{ln(2)}right)^2+frac{lnleft(x-frac{5}{4}right)}{ln(2)}right)ln(x)=frac{1}{2}ln(2)$$ and here you can take a numerical method
with such a method we obtain $$xapprox 2.050105808$$
answered Apr 14 '17 at 14:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
$begingroup$
this is turning out to be a tricky thing. can u please solve it?
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:58
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
taking the logarithm on both sides of your equation we obatin
$$left(frac{3}{4}left(frac{ln(x)}{ln(2)}right)^2+frac{lnleft(x-frac{5}{4}right)}{ln(2)}right)ln(x)=frac{1}{2}ln(2)$$ and here you can take a numerical method
with such a method we obtain $$xapprox 2.050105808$$
answered Apr 14 '17 at 14:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
$begingroup$
this is turning out to be a tricky thing. can u please solve it?
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:58
add a comment |
$begingroup$
taking the logarithm on both sides of your equation we obatin
$$left(frac{3}{4}left(frac{ln(x)}{ln(2)}right)^2+frac{lnleft(x-frac{5}{4}right)}{ln(2)}right)ln(x)=frac{1}{2}ln(2)$$ and here you can take a numerical method
with such a method we obtain $$xapprox 2.050105808$$
answered Apr 14 '17 at 14:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
$begingroup$
this is turning out to be a tricky thing. can u please solve it?
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:58
add a comment |
$begingroup$
taking the logarithm on both sides of your equation we obatin
$$left(frac{3}{4}left(frac{ln(x)}{ln(2)}right)^2+frac{lnleft(x-frac{5}{4}right)}{ln(2)}right)ln(x)=frac{1}{2}ln(2)$$ and here you can take a numerical method
with such a method we obtain $$xapprox 2.050105808$$
answered Apr 14 '17 at 14:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
taking the logarithm on both sides of your equation we obatin
$$left(frac{3}{4}left(frac{ln(x)}{ln(2)}right)^2+frac{lnleft(x-frac{5}{4}right)}{ln(2)}right)ln(x)=frac{1}{2}ln(2)$$ and here you can take a numerical method
with such a method we obtain $$xapprox 2.050105808$$
answered Apr 14 '17 at 14:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
edited Apr 14 '17 at 15:24
answered Apr 14 '17 at 14:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
answered Apr 14 '17 at 14:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
answered Apr 14 '17 at 14:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
76.8k42866
$begingroup$
this is turning out to be a tricky thing. can u please solve it?
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:58
add a comment |
$begingroup$
this is turning out to be a tricky thing. can u please solve it?
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:58
$begingroup$
this is turning out to be a tricky thing. can u please solve it?
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:58
$begingroup$
this is turning out to be a tricky thing. can u please solve it?
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:58
add a comment |
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$begingroup$
Does $log^2 x$ mean $log(log x)$ or $(log x)^2$
$endgroup$
– DHMO
Apr 14 '17 at 14:39
1
$begingroup$
The title and the question have different equations in them. Which one do you mean?
$endgroup$
– AsafHaas
Apr 14 '17 at 14:40
$begingroup$
$ (log_2 x) ^ 2 $
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
@AsafHaas its the same. i hav copied it from the title.
$endgroup$
– Esha Mukhopadhyay
Apr 14 '17 at 14:41
$begingroup$
By executing your idea and let $y=log_2 x$, we get $1/2=y(3y^2/4+log_2{(x-5/4)})$
$endgroup$
– didgogns
Apr 14 '17 at 14:43