How do it prove this without approximately calculating the logarithm ?

Question

asked Mar 1, 2024 176k views

1 Answer

Timothy Brown · Answer 1 · 2024-03-08T10:51:54+0000

Answer:

$10 < {e}^(8)$

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