# Verify if` log_3 5 + log_5 3 gt 2`

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### 2 Answers

You need to convert the given logarithms in natural logarithms, such that:

`(ln 5)/(ln 3) + (ln 3)/(ln 5) > 2`

Bringing the terms to the left side to a common denominator yields:

`(ln^2 5 + ln^2 3)/(ln 3*ln 5) > 2`

You may replace `(ln 5 + ln 3)^2 - 2ln5*ln 3` for` ln^2 5 + ln^2 3` , such that:

`((ln 5 + ln 3)^2 - 2ln5*ln 3)/(ln 3*ln 5) > 2`

`((ln 5 + ln 3)^2)/(ln 3*ln 5) - 2 > 2`

`((ln 5 + ln 3)^2) > 4(ln 3*ln 5)`

Expanding the square yields:

`ln^2 5 + 2(ln 3*ln 5) + ln^2 3 > 4(ln 3*ln 5)`

`ln^2 5 + 2(ln 3*ln 5) -4(ln 3*ln 5) + ln^2 3 > 0`

`ln^2 5 - 2(ln 3*ln 5) + ln^2 3 > 0`

`(ln 5 - ln 3)^2 > 0` valid

**Hence, testing if the inequality `log_3 5 + log_5 3 > 2` holds yields that `(ln 5 - ln 3)^2 > 0` , hence `log_3 5 + log_5 3 > 2` holds.**

We'll re-write the inequality:

log 3 5 + 1/log 3 5 > 2

We'll note log 3 5 =a

We'll re-write the inequality in a:

a + 1/a > 2

We'll multiply by 2 the inequality:

a^2 + 1 > 2a

We'll subtract 2a both sides:

a^2 - 2a + 1 > 0

(a - 1)^2 > 0

The inequality is true for any value of a.

(log 3 5 - 1)^2 > 0