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| bio | website | |
|---|---|---|
| location | Kolkata, India | |
| age | ||
| visits | member for | 1 year |
| seen | 11 mins ago | |
| stats | profile views | 2,103 |
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Jun 16 |
revised |
Euler-Fermat Theorem added 385 characters in body |
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Jun 16 |
answered | Euler-Fermat Theorem |
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Jun 16 |
comment |
$a^{\log_b(c)} = c^{\log_b(a)}$ @user82627, As $$\log_bm=\frac{\log m}{\log b}, \log _ea\cdot \log_bc=\frac{\log a}{\log e}\cdot \frac{\log c}{\log b}=\frac{\log c}{\log e}\cdot \frac{\log a}{\log b}=\log_ec\cdot\log_ba $$ |
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Jun 16 |
awarded | algebra-precalculus |
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Jun 15 |
awarded | Nice Answer |
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Jun 15 |
answered | $a^{\log_b(c)} = c^{\log_b(a)}$ |
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Jun 15 |
comment |
Can you explain me this antiderivative? @João, what Wolfram presents? |
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Jun 15 |
revised |
How to find a solution(s) when given two equations of two variables? added 201 characters in body |
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Jun 15 |
answered | How to find a solution(s) when given two equations of two variables? |
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Jun 15 |
answered | Exponential function passing through two points. |
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Jun 15 |
answered | Inverse of the function |
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Jun 15 |
comment |
$3^{3n+1} < 2^{5n+6} $ for all non-negative integers $n$. Is my induction solution correct? If $$2^{5k+6}>3^{3k+1},$$ $$2^{5(k+1)+6}=2^5\cdot2^{5k+6} >32\cdot3^{3k+1}>3^3\cdot3^{3k+1}(\text{ as } 32>27)=3^{3(k+1)+1} $$ |
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Jun 15 |
comment |
How to calculate a bound for this product? How about applying logarithm & then expand using $\ln(1-x)$ |
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Jun 15 |
revised |
Modular arithmetic: How to solve $3^{n+1} \equiv 1 \pmod{11}$? added 534 characters in body |
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Jun 15 |
answered | Evaulate $\lim_{x\to\infty}\frac{3x^2-36x+12}{5x^2+113x-2}$ |
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Jun 15 |
comment |
Prove $n\mid \phi(2^n-1)$ @AbhraAbirKundu, may be my comment & your edit were concurrent. Anyway, good job. |
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Jun 15 |
comment |
Prove $n\mid \phi(2^n-1)$ @Abhra, this can be generalized to $n|\phi(a^n-1)$ for any integer $a$ |
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Jun 15 |
revised |
Modular arithmetic: How to solve $3^{n+1} \equiv 1 \pmod{11}$? added 199 characters in body |
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Jun 15 |
answered | Modular arithmetic: How to solve $3^{n+1} \equiv 1 \pmod{11}$? |
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Jun 15 |
revised |
Fibonacci sequence: how to prove that $\alpha^n=\alpha\cdot F_n + F_{n-1}$? deleted 29 characters in body |
