What are the rules of powers of powers? [duplicate]

This question already has answers here: Incorrect notation in math? (4 answers) Closed 6 years ago. What would 234 equate to? I can think of two rules that may apply: abc=a(bc) (Making 234=281≈2.417⋅1024) or abc=(ab)c=abc (Making 234=212=4096) Which of these is true? Answer As Dietrich Burde stated in a comment, the standard convention is that … Read more

Find all xx such that 2x,2×22^x,2^{x^2} and 2×32^{x^3} form 33 terms of an A.P.

I know that if a,b,c are in Arithmetic Progression, then 2b=a+c, but in this case, I am unable to solve for x. Hints are appreciated. Thanks. Answer HINT: 2x(1+2×3−x)=2×2+1⟺1+2×3−x=2×2−x+1 If x3−x>0,1+2×3−x>1 is odd unlike 2×2−x+1 as x2−x+1>0 for real x So, x3−x must be 0 and consequently, 2=2×2−x+1⟹1=x2−x+1 AttributionSource : Link , Question Author : … Read more

Need assistance in solving exponential equation: 27x92x−1=3x+4\frac{27^x}{9^{2x-1}}=3^{x+4}

Find value of x: 27x92x−1=3x+4 My steps: (33)x(32)2x−1=3x+4 3x4x−2=x+4 Please help me finish solving, and correct me if what I did so far has mistakes. Thanks very much. Answer Your last step doesn’t work, instead do: 33x−(4x−2)=3x+4. AttributionSource : Link , Question Author : Arthur Alex Karapetov , Answer Author : Thomas

Finding lim\lim_{n \to \infty} \dfrac{n^n}{(2n)!}

Struggling to apply Squeeze THM to find this limit. Specifically, I need a sequence which is always larger than the one in the problem, but which can easily be derived from the middle sequence. Answer In the denominator (2n)!= (2n)(2n-1) \dots (n+1)\cdot n!. Each of the factors from (n+1) to (2n) are larger than n; … Read more

\lim a^{b_n}=a^b\lim a^{b_n}=a^b if \lim b_n = b\lim b_n = b [duplicate]

This question already has answers here: how to prove that \lim x_n^{y_n}=\lim x_n^{\lim y_n}? (2 answers) Closed 5 years ago. Let a \in \mathbb{R} and \lim b_n = b (for a real sequence b_n). I am trying to show that \lim a^{b_n}=a^b by using the definition of the convergence of a sequence, but I cannot … Read more

Evaluate this without a calculator

Evaluate 12017+22017+…+10002017(mod2016) These large exponents prevent me from finding any quick method to find the mod. Answer 2016=25⋅32⋅7, so do it mod 25, 32 and 7 and use Chinese Remainder Theorem. For 25=32, note that ϕ(25)=16, and 2017 \equiv 1 \mod 16 so a^{2017} \equiv a \mod 2^5 if a is odd, while a^{2017} \equiv … Read more