log question

Question

if (x lg x)=4608 , then how to calculate value of x ?

Base is 2

Answer 1

Just try substituting, $9 . 2^9 = 4608, x = 512$

Comments

I think there should be some method to solve it .

I reached upto 

x lg x =4608

lg x^x = 4608

x^x = 2 ⁴⁶⁰⁸

Approximately  x!= 2⁴⁶⁰⁸

^{How to proceed after this , or any other method than substitution .}

^{Anyways , thanks for Answer :-)}

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@Rohan: You can't be guaranteed that the number will behave nicely when you try to manipulate it algebraically.

Hit and trial is a much better way to do it, especially since you can use binary search.

If I were to do it on paper, I would make an educated guess, then improve it iteratively (using Binary search in my head perhaps since it is so easy to perform mentally). If I had access to a computing device, I would implement a numerical technique like Newton Raphson to converge on the solution faster.

My mental process:
$x = 1024, \log_2 x = 10$ ... Too big.
$x = 256, \log_2 x = 7$... Too small.
$x = 512, \log_2 x = 9$... $512 \times 9 = 4608$. Bingo!

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@Rohan:

Also, this step is incorrect.

$x^x=2^{4608}$

Approximately  $x!=2^{4608}$

Note that $x\log{x} = \Theta(\log{x!})$, but $x^x \neq \Theta(x!)$

Infact, $x!$ grows much slower than $x^x$

For example,

$512^{512} = 2^{4608}$, but $512! \approx 2^{3874}$. This is incorrect by a factor of $10^{220}$ (which is extremely incorrect!)

Answer 2

We factor $4608$ and obtain =$512 \times 9$

also from the equation on rearrangement:
$\begin{align*} x\log_2{x} &= 4608 \\ \log_2{x^x} &= 4608 \\ x^x &= 2^{4608}\\ &= 2^{512\times 9}\\ &= \Big(2^{512}\Big)^{9}\\ &= \Big(2^{9}\Big)^{512}\\ &= \Big(2^{9}\Big)^{2^9}\\ x^x &= 512^{512} \end{align*}$

We get $x = 512$

Comments

You kinda got lucky here because 4608 factored nicely.

If the number had been something like 4000, you would've been stuck with this method.

$x^x = 2^{4000} = (2^?)^{2^?}$

Hit and trail is definitely the way to go. A better way is to binary search for the answer.

PS: look at my comments on Arjun sir's answer.