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secondary 2 | Maths
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logarithms
Date Posted: 3 years ago
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logx/9 (x²/3) < 6 + log3(9/x)
log3 (x²/3) / log3 (x/9) < 6 + log3(9) - log3(x)
(log3(x²) - log3(3) ) / (log3(x) - log3(9)) < 6 + log3(3²) - log3(x)
(2log3(x) - 1) / (log3(x) - log3(3²)) < 6 + 2 - log3(x)
(2log3(x) - 1) / (log3(x) - 2) < 8 - log3(x)
Use the substitution u = log3(x),
(2u - 1) / (u - 2) < 8 - u
(2u - 1) / (u - 2) + u - 8 < 0
(2u - 1 + (u - 8)(u - 2)) / (u-2) < 0
(2u - 1 + u² - 2u - 8u + 16)/(u - 2) < 0
(u² - 8u + 15) / (u - 2) < 0
(u - 5)(u - 3) / (u - 2) < 0
Substitute values whereby u > 5, 3 < u < 5 , 2 < u < 3 and u < 2 to test for the sign of the expression.
When (u - 5)(u - 3) / (u - 2) < 0
3 < u < 5 and u < 2
So,
3 < log3(x) < 5 and log3(x) < 2
log3(3³) < log3(x) < log3(3^5) and log3(x) < log3(3²)
3³ < x < 3^5 and x < 3²
27 < x < 243 and x < 9
But x has to be > 0 for the logarithm to be defined so 0 < x < 9
Number of positive integers satisfying 27 < x < 243
= 243 - 26 - 2
= 215
(243 - 26 is the number of integers from 27 to 243 inclusive. Then - 2 since we can't count 27 and 243)
Number of positive integers satisfying 0 < x < 9
Number of integers from 1 to 8 = 8
Total = 215 + 8 = 223
log3 (x²/3) / log3 (x/9) < 6 + log3(9) - log3(x)
(log3(x²) - log3(3) ) / (log3(x) - log3(9)) < 6 + log3(3²) - log3(x)
(2log3(x) - 1) / (log3(x) - log3(3²)) < 6 + 2 - log3(x)
(2log3(x) - 1) / (log3(x) - 2) < 8 - log3(x)
Use the substitution u = log3(x),
(2u - 1) / (u - 2) < 8 - u
(2u - 1) / (u - 2) + u - 8 < 0
(2u - 1 + (u - 8)(u - 2)) / (u-2) < 0
(2u - 1 + u² - 2u - 8u + 16)/(u - 2) < 0
(u² - 8u + 15) / (u - 2) < 0
(u - 5)(u - 3) / (u - 2) < 0
Substitute values whereby u > 5, 3 < u < 5 , 2 < u < 3 and u < 2 to test for the sign of the expression.
When (u - 5)(u - 3) / (u - 2) < 0
3 < u < 5 and u < 2
So,
3 < log3(x) < 5 and log3(x) < 2
log3(3³) < log3(x) < log3(3^5) and log3(x) < log3(3²)
3³ < x < 3^5 and x < 3²
27 < x < 243 and x < 9
But x has to be > 0 for the logarithm to be defined so 0 < x < 9
Number of positive integers satisfying 27 < x < 243
= 243 - 26 - 2
= 215
(243 - 26 is the number of integers from 27 to 243 inclusive. Then - 2 since we can't count 27 and 243)
Number of positive integers satisfying 0 < x < 9
Number of integers from 1 to 8 = 8
Total = 215 + 8 = 223
Edited
See 1 Answer
Mistake from 1 of the steps onwards. You can't multiply by (y - 2) as it is not known if y - 2 is positive or negative.
It is an inequality sign and not an equals sign. You can only multiply by y - 2 if you know what y-2 's sign is. (and therefore decide whether to flip the sign or not)
Or,
when it's an equals sign.
The terms on the right hand side should be brought to the left , then factorised. Sign checking is done afterwards.
You may refer to my working on the main comments section.
It is an inequality sign and not an equals sign. You can only multiply by y - 2 if you know what y-2 's sign is. (and therefore decide whether to flip the sign or not)
Or,
when it's an equals sign.
The terms on the right hand side should be brought to the left , then factorised. Sign checking is done afterwards.
You may refer to my working on the main comments section.
Secondly, log3(x) = 4 is incorrect also.
the inequality is found to be 3 < y < 5. Replacing y with log3(x) gives :
3 < log3(x) < 5
Question gives x to be positive integer. We have to first solve this inequality before deciding which positive values of x satisfy it.
i.e
log3(3³) < log3(x) < log3(3^5)
3³ < x < 3^5
27 < x < 243
So the possible values of x are positive integers from 28 to 242, inclusive.
The other inequality that you've missed out was 0 < log3(x) < 2
the inequality is found to be 3 < y < 5. Replacing y with log3(x) gives :
3 < log3(x) < 5
Question gives x to be positive integer. We have to first solve this inequality before deciding which positive values of x satisfy it.
i.e
log3(3³) < log3(x) < log3(3^5)
3³ < x < 3^5
27 < x < 243
So the possible values of x are positive integers from 28 to 242, inclusive.
The other inequality that you've missed out was 0 < log3(x) < 2
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