How many positive integers x satisfy this logarithmic inequality?

[Mr X]

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.

 

[PeroK]

Did you try plugging in $x = 1, 2, 3 \dots$ and see what happens?

 

[pasmith]

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 $$\log_{x/8}(x^2/4) = 2\frac{ \log_2(x/4)}{\log_2(x/8)} = \frac{2 (\log_2 x - 1)}{\log_2 x - 3}$$ and the right hand side is $$7 + \log_2(8/x) = 10 - \log_2 x$$ so that if $y = \log_2 x$ then $$\frac{2(y-1)}{y-3} < 10 - y.$$

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

Alternatively, if $y < 3$ then we must reverse the inequality when multiplying both sides by $y - 3$, which leads to $$(y - 4)(y - 7) > 0.$$

 

[Mr 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 $$\log_{x/8}(x^2/4) = 2\frac{ \log_2(x/4)}{\log_2(x/8)} = \frac{2 (\log_2 x - 1)}{\log_2 x - 3}$$ and the right hand side is $$7 + \log_2(8/x) = 10 - \log_2 x$$ so that if $y = \log_2 x$ then $$\frac{2(y-1)}{y-3} < 10 - y.$$

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

Alternatively, if $y < 3$ then we must reverse the inequality when multiplying both sides by $y - 3$, which leads to $$(y - 4)(y - 7) > 0.$$

sorry I didn't reply after, but got it, thankyou.