How many positive integers x satisfy this logarithmic inequality?

  • #1
Mr X
24
3
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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  • #2
Did you try plugging in ##x = 1, 2, 3 \dots## and see what happens?
 
  • #3
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 [tex]
\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}[/tex] and the right hand side is [tex]7 + \log_2(8/x) = 10 - \log_2 x[/tex] so that if [itex]y = \log_2 x[/itex] then [tex]
\frac{2(y-1)}{y-3} < 10 - y.[/tex]

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

Alternatively, if [itex]y < 3[/itex] then we must reverse the inequality when multiplying both sides by [itex]y - 3[/itex], which leads to [tex](y - 4)(y - 7) > 0.[/tex]
 
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Likes Mr X and nuuskur
  • #4
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 [tex]
\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}[/tex] and the right hand side is [tex]7 + \log_2(8/x) = 10 - \log_2 x[/tex] so that if [itex]y = \log_2 x[/itex] then [tex]
\frac{2(y-1)}{y-3} < 10 - y.[/tex]

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

Alternatively, if [itex]y < 3[/itex] then we must reverse the inequality when multiplying both sides by [itex]y - 3[/itex], which leads to [tex](y - 4)(y - 7) > 0.[/tex]
sorry I didn't reply after, but got it, thankyou.
 

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