How do you solve $e^{x} > 1.6$ $e^x>1.6$ ?

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How do you solve $e^{x} > 1.6$ $e^x>1.6$ ?

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Answer 1 · 2017-02-07T21:56:34

$x > ln 1.6$ $x > ln1.6$ , with $ln 1.6 \approx 0.470003629$ $ln1.6 ~~ 0.470003629$ .

Explanation:

Since $ln x$ $lnx$ is a constantly increasing function, for any $a > b$ $a > b$ it holds that $ln a > ln b$ $lna > lnb$ . Applying this to the relationship:

${ln e}^{x} > ln 1.6$ $ln e^x > ln1.6$ , and we know that ${ln e}^{x} = x$ $ln e^x = x$

If even further confirmation is required, from logarithm properties:

${ln e}^{x} = x ln e = x \cdot 1 = x$ $ln e^x = xlne = x* 1 = x$

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How do you solve $e^{x} > 1.6$ $e^x>1.6$ ?

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