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

First step:

$$3^1 < {\log_{2}{x}} < 3^2$$

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:

$$2^{3^{1}}<x<2^{3^{2}}$$
=>$$2^{3}<x<2^{9}$$

 

[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:

$$2^{3}<x<2^{9}$$

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]

$$1 < \log_3({\log_{2}{x}}) < 2$$

$$3^1 < {\log_{2}{x}} < 3^2$$

$$3 < {\log_{2}{x}} < 9$$

$$2^3 < x < 2^9$$

$$8 < x < 512$$

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. :)