How many positive integers x satisfy this logarithmic inequality?

AI Thread Summary
The discussion revolves around solving the logarithmic inequality involving positive integers x. The key point is the transformation of the inequality into a quadratic form after substituting y = log₂ x. The solution requires careful handling of the inequality when multiplying by (y - 3), as it affects the direction of the inequality depending on the sign of (y - 3). The final factorization leads to conditions for y that determine the valid range for x. Ultimately, the participants clarify the steps necessary to find the number of positive integer solutions for x.
Mr X
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Homework Statement
How many positive integers x satisfy ##\log_{\frac x 8} (\frac{x^2} 4) < 7 + \log_2(\frac 8 x)##
Relevant Equations
Basic logarithamic rules
The whole solution is a bit long, which I'll attach but the part I'm stuck at is, assuming everything else above it is correct, is
4 < (log x - 3)(8-log x)

Note ; inequalities aren't technically taught yet in the course, so please try to make the solution not go too deep into that. If that isn't possible there's a high chance I've gone wrong before that step, hence I've uploaded the entire solution.
 

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Did you try plugging in ##x = 1, 2, 3 \dots## and see what happens?
 
Mr X said:
Homework Statement: How many positive integers x satisfy ##\log_{\frac x 8} (\frac{x^2} 4) < 7 + \log_2(\frac 8 x)##
Relevant Equations: Basic logarithamic rules

The whole solution is a bit long, which I'll attach but the part I'm stuck at is, assuming everything else above it is correct, is
4 < (log x - 3)(8-log x)

Assuming your logs are to base 2 then this is one of the possible solutions. It is convenient to use logs to base 2; then the left hand side is <br /> \log_{x/8}(x^2/4) = 2\frac{ \log_2(x/4)}{\log_2(x/8)} = \frac{2 (\log_2 x - 1)}{\log_2 x - 3} and the right hand side is 7 + \log_2(8/x) = 10 - \log_2 x so that if y = \log_2 x then <br /> \frac{2(y-1)}{y-3} &lt; 10 - y.

To solve this, we must multiply both sides by y - 3. But this only preserves the inequality if y - 3 &gt; 0, and in that case <br /> y^2 - 11y + 28 &lt; 0 which is as far as you got, but you then did not factorise this as <br /> (y - 4)(y - 7) &lt; 0.

Alternatively, if y &lt; 3 then we must reverse the inequality when multiplying both sides by y - 3, which leads to (y - 4)(y - 7) &gt; 0.
 
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pasmith said:
Assuming your logs are to base 2 then this is one of the possible solutions. It is convenient to use logs to base 2; then the left hand side is <br /> \log_{x/8}(x^2/4) = 2\frac{ \log_2(x/4)}{\log_2(x/8)} = \frac{2 (\log_2 x - 1)}{\log_2 x - 3} and the right hand side is 7 + \log_2(8/x) = 10 - \log_2 x so that if y = \log_2 x then <br /> \frac{2(y-1)}{y-3} &lt; 10 - y.

To solve this, we must multiply both sides by y - 3. But this only preserves the inequality if y - 3 &gt; 0, and in that case <br /> y^2 - 11y + 28 &lt; 0 which is as far as you got, but you then did not factorise this as <br /> (y - 4)(y - 7) &lt; 0.

Alternatively, if y &lt; 3 then we must reverse the inequality when multiplying both sides by y - 3, which leads to (y - 4)(y - 7) &gt; 0.
sorry I didn't reply after, but got it, thankyou.
 
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