Number of Integers Satisfying 1<log₃(log₂x)<2

[yik-boh]

How many integers will satisfy x in the inequality:

1< \log_{3}({\log_{2}{x})}< 2

Note: The log there is not multitplied to the other log. The log there I think is read like this, logarithm of logarithm of x to the base 2 to the base 3.

What can be the solution or technique for this one? This was given on a math contest here and was just asked to solve for 20 seconds.

 

[g_edgar]

How many integers will satisfy x in the inequality:
1&lt; \log_{3}({\log_{2}{x})} &lt; 2

First step:

3^1 &lt; {\log_{2}{x}} &lt; 3^2

Can you find the second step?

 

[yik-boh]

I forgot to mention the answer. It's 503.

Trying the step you gave me:

2^{3}=8 < x < 2^{6}=64

so the new equation would be like this

8 < x < 64

After that, I multiplied 64 to 8. I got 512.

When I got 512, I subtracted 3^{2} from 512 then I got 503.

Is my method correct?

 

[icystrike]

Why did you multiply 64 to 8 ?

My method would be:

2^{3^{1}}&lt;x&lt;2^{3^{2}}
=>2^{3}&lt;x&lt;2^{9}

 

[yik-boh]

When you raise an exponent to another exponent, you should multiply the exponents right? So 2^{3^{2}} would be 2^{6}.

Please explain to me step by step what to do. I'm still confused. I'm just new to this type of problems.

 

[zgozvrm]

When you raise an exponent to another exponent, you should multiply the exponents right? So 2^{3^{2}} would be 2^{6}.

Please explain to me step by step what to do. I'm still confused. I'm just new to this type of problems.

No. For example, 3^2 = 9 and 2^{(3^2)} = 2 ^ 9 = 512 but 2^6 = 64

Therefore, a^{b^c} \neq a^{bc}

 

[yik-boh]

Oh thanks for the explanation dude..

So what would I do next after this:

2^{3}&lt;x&lt;2^{9}

to get 503?


Hope you could explain it step by step. Thanks. :)

 

[Ben1220]

You could just think about it for a minute :P

Your asking how many integers fall between an interval...

 

[yik-boh]

Please correct me if I'm wrong.

After arriving at this one:

8 < x < 512

I transposed 8 to the side of 512 so the equation would be:

x < 512 - 8

x < 504

The number just before 504 is 503. So the answer is, there are 503 possible values for x in order to satisfy the inequality. Which is the right answer.But is my solution and reasoning correct? :)

 

[g_edgar]

1 &lt; \log_3({\log_{2}{x}}) &lt; 2

3^1 &lt; {\log_{2}{x}} &lt; 3^2

3 &lt; {\log_{2}{x}} &lt; 9

2^3 &lt; x &lt; 2^9

8 &lt; x &lt; 512

answer 512-8-1=503

 

[HallsofIvy]

When you raise an exponent to another exponent, you should multiply the exponents right? So 2^{3^{2}} would be 2^{6}.

Please explain to me step by step what to do. I'm still confused. I'm just new to this type of problems.

(a^ b)^c= a^{bc} but a^{b^c} is not.

 

[Ben1220]

Please correct me if I'm wrong.

After arriving at this one:

8 < x < 512

I transposed 8 to the side of 512 so the equation would be:

x < 512 - 8

x < 504

The number just before 504 is 503. So the answer is, there are 503 possible values for x in order to satisfy the inequality. Which is the right answer.


But is my solution and reasoning correct? :)

yes.

 

[yik-boh]

Thanks for the help. :)