Question

How do it prove this without approximately calculating the logarithm ?​

Answer 1

`10 < {e}^(8)`

`ln(10) < 8`

`ln(10) - 8 < 0`

Since ln(10) > 0, it follows that ln(10) + 1 > 0, and so:

`( ln(10) + 1)( ln(10) - 8) < 0`

`{( ln(10)) }^(2) - 7 ln(10) - 8 < 0`

`( { ln(10) })^(2) - 7 ln(10) + 1 < - 9`

`{( ln(10)) }^(2) + 2 ln(10) + 1 < 9 ln(10) - 9`

`{( ln(10) )}^(2) + 2 ln(10) + 1 < 9( ln(10) - 1)`

(ln(10) + 1)^2 < 9(ln(10 - 1)

ln(10) + 1 < 3√(ln(10) - 1)

(1 + ln(10))/3 < √(ln(10) - 1)

√(ln(10) - 1) > (1+ ln(10))/3