Since 1 / log₃x = logₓ3 , the series can be rewritten as :

logₓ3 + logₓ3³ + logₓ3⁵ + ... + logₓ3²ⁿ⁻¹


Sum of the first 30 terms

= logₓ3 + logₓ3³ + logₓ3⁵ + ... + logₓ3²⁽³⁰⁾⁻¹

= logₓ3 + 3logₓ3 + 5logₓ3 + ... + 59logₓ3

= (1 + 3 + 5 + ... + 59) logₓ3

The terms in the brackets form an arithmetic progression with first term, a = 1 and common difference, d = 2 and last term, l = 59.

Using the sum of A.P formula Sn = ½n(a + l) ,

= (½(30)(1 + 59)) logₓ3

= 900 logₓ3


So,

900 logₓ3 ≥ 300

3 logₓ3 ≥ 1

3 ln3 / ln x ≥ 1

( Recall logₐb = log꜀b / log꜀a )

b)

3ln3 / lnx ≥ 1

3ln3 / lnx - 1 ≥ 0

(3ln3 - lnx) / lnx ≥ 0

(ln3³ - lnx) / ln x ≥ 0

ln(27/x) / ln x ≥ 0


Now lnx = 0 when x = 1. This makes the denominator equal 0 and the expression becomes undefined.

We also know that when x = 27, the numerator ln(27/x) = ln1 = 0 . The denominator lnx is non-zero and positive.

This makes the entire expression equal 0 (0 divided by a non-zero value is still 0) , which satisfies the inequality.


And we know x cannot be ≤ 0 since lnx and ln(27/x) would be undefined as well.


Since we want ln(27/x) / ln x ≥ 0, we can test for values of x in the domains 0 < x < 1, 1 < x < 27 and x > 27

(x = 27 has already been shown above)


For 0 < x < 1,

27/x > 1, so ln(27/x) > 0, but lnx < 0 .

With a positive numerator and negative denominator, the expression < 0


For 1 < x < 27 ,

27/x > 1 so ln(27/x) > 0, and lnx > 0 as well.
With both numerator and denominator being positive, the expression > 0


For x > 27,

27/x < 1 , so ln(27/x) < 0 , but lnx > 0.
With a negative numerator and positive denominator, the expression < 0


So the range of values that satusfies the inequality is :


1 < x ≤ 27