Tasty Logarithm

Algebra Level 3

$\large \log_2(\log_{2^x}(\log_{2^y}(2^{1000}))) = 0$

If $x$ and $y$ are positive integers satisfying the equation above, then find the sum of all possible values of $x+y$ .

The answer is 881.

2 solutions

Brian Charlesworth
May 9, 2015

We require that

$\large\log_{2^{x}}(\log_{2^{y}}(2^{1000})) = 1 \Longrightarrow \log_{2^{y}}(2^{1000}) = 2^{x}$

$\large \Longrightarrow (2^{y})^{2^{x}} = 2^{1000} \Longrightarrow y \times 2^{x} = 1000.$

Now since $1000 = 2^{3} \times 5^{3}$ we can have $x = 1,2,3,$ which corresponds to respective values for $y$ of $500, 250, 125.$

Thus the desired sum of possible values for $x + y$ is $501 + 252 + 128 = \boxed{881}.$

Moderator note:

Yes, the crux of this problem is simply applying the properties of indices correctly: $(2^a)^b = 2^{a \times b}$ . Nicely done.

Bonus question: with another unknown positive integer $z$ . Find the value of $x+y+z$ such that the equation below is fulfilled.

$\large \log_2(\log_{2^x}(\log_{2^y} (\log_{2^z} (2^{1000})))) = 0$

Re Bonus question: Using a similar method, we end up with the equation

$\large z \times 2^{y*2^{x}} = 1000.$

Given the prime factorization of $1000$ , and the fact that $x,y \ge 1,$ we can only have $y*2^{x} = 1*2^{1} = 2,$ which makes $z = 2 \times 5^{3} = 250.$

Thus $x + y + z = 1 + 1 + 250 = 252.$

Brian Charlesworth - 6 years, 1 month ago

for all users who are saying that x=0 can also be include in solution, there is a line in question that x and y are positive integers. and also x=0 can't give value while taking log

Dhirendra Singh - 6 years, 1 month ago

Good work.followed same procedure as yours.

shivamani patil - 6 years ago

i am stop here x*y=500 and 252 not correct answer

Patience Patience - 5 years, 1 month ago

I think ans should be 4680

Nilesh Rao - 4 years, 11 months ago

By solving I am getting x*y=1000

Nilesh Rao - 4 years, 11 months ago

Answer has to be 1881, why can't we take x=0 case. Is " 0 " not considered as positive integer.?

Kiran Bk - 6 years, 1 month ago

If we try x=0, then we take $log_1 (log_{2^y} (2^{1000}))$ . $log_1 x$ cannot return a number because if $log_1 x = y$ , then $1^y=x$ which only works if x=1. Also, y could be an infinite number of cases. We see here that there would be no argument returned and thus the answer would not be 0.

Sean Engelstad - 6 years, 1 month ago

No 0 is not a positive neither a negative no.

Shashanka Indri - 6 years, 1 month ago

0 is a base element to decide if a number is negative or positive. It is neither negative nor positive integer.

Dhirendra Singh - 6 years, 1 month ago

Popular Power
Jun 9, 2019

The problem simplifies to:

$\log_{2^y}{2^{1000}}=2^x \Rightarrow \dfrac{1000}{y}=2^x$

Remember that $y$ and $2^x$ has to be a factor of $1000$ . So, we have $1000=2^3\times5^3$ . So, $x=1,2,3$ satisfy the given equation. After gaining the subsequent values of $y$ as $500,250,125$ , we just add them all up.

$x+y=500+250+125+3+2+1=881$

Tasty Logarithm

The answer is 881.

2 solutions

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