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Mean
Mode
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Mixed Fractions
Prime Factorization
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Combine Like Terms
Solve for a Variable
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Evaluate
\ln(2)\approx 0.693147181
Quiz
Algebra
5 problems similar to:
\log _ { e } 2
Similar Problems from Web Search
What is \displaystyle{{\log}_{{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 \displaystyle{{\log}_{{2}}{\left({e}\right)}} ?
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: \displaystyle{{\log}_{{a}}{a}}={1} , \displaystyle{{\log}_{{e}}{e}}={\left({{\log}_{{2}}{e}}\right)}{\left({{\log}_{{e}}{2}}\right)} ...
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 ...
Transformation of exponentials
https://math.stackexchange.com/questions/1598335/transformation-of-exponentials
Both of your answers are correct as I pointed out in my comment:\log_3 e=\frac{\log_e e}{\log_e 3}=\frac{1}{\log_e 3}
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^{\log_b(x)} = x
https://math.stackexchange.com/questions/2190057/trouble-understanding-when-to-use-b-log-bx-x
You can always write x=b^{\log_bx} when x,b>0. So that means that if you felt like it you could indeed write x=b^{\log_bx}=b^{\log_bb^{\log_bx}}=b^{\log_bb^{\log_bb^{\log_bx}}}=\ldots ...
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Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}
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