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

How do it prove this without approximately calculating the logarithm ?​-example-1

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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

User Timothy Brown
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