Solve log_{e}2 | Microsoft Math Solver

Evaluate

ln (2) \approx 0.693147181

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5 problems similar to:

lo g_{e} 2

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lo g_{e} 5

https://socratic.org/questions/what-is-log-e-5

Log_e (5) = 1.609 Explanation: Log_e (5) = ln(5) = 1.609 where n stands for natural logarithm. The natural logarithm is such that ln(e) =1. Alternatively, Log_e(5) is the number such that e^x = 5.

How do you differentiate

lo g_{2} (e)

https://socratic.org/questions/how-do-you-differentiate-log-2-e

This is a constant. The derivative of any constant, with respect to any variable, is 0. Explanation:

lo g_{a} a = 1

lo g_{e} e = (lo g_{2} e) (lo g_{e} 2)

Stuck on a problem about the length of a partial sum

https://math.stackexchange.com/questions/2827117/stuck-on-a-problem-about-the-length-of-a-partial-sum

The alternating series theorem gives you a bound. For a series where the signs alternate and the terms decrease in magnitude the error is always of the sign of the first neglected term and smaller ...

Is this approach correct? I can't arrange my result, so that it's equal to the solution, is there any mistake? Power series.

https://math.stackexchange.com/questions/2075236/is-this-approach-correct-i-cant-arrange-my-result-so-that-its-equal-to-the-s

Could make it simpler: \begin{align*} \ln (a+bz) &= \ln \left[ a\left(1+\frac{bz}{a} \right) \right] \\ &= \ln a+\ln \left(1+\frac{bz}{a} \right) \\ &= \ln a+\sum_{n=1}^{\infty} ...

trouble understanding when to use

b^{l o g_{b} (x)} = x

https://math.stackexchange.com/questions/2190057/trouble-understanding-when-to-use-b-log-bx-x

You can always write

x = b^{l o g_{b} x}

x, b > 0

. So that means that if you felt like it you could indeed write

x = b^{l o g_{b} x} = b^{l o g_{b} b^{l o g_{b} x}} = b^{l o g_{b} b^{l o g_{b} b^{l o g_{b} x}}} = \dots

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