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Solution

Let $a=\log_2(5·2^x+1)$, $b=\log_4(2^{1-x}+1)$, $c=1$ For A.P.: $2b=a+c$ Simplify using $\log_4 y = \frac{1}{2}\log_2 y$ After algebra → $x = 1 - \log_2 5$

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Solution

For any integer $n\ge1$, digits $= \lfloor\log_{10} n\rfloor+1$.

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Solution

A.P. $\Rightarrow 2\cdot\log_{9}(3^{1-x}+2)=1+\log_{3}(4\cdot 3^x-1)$. Since $\log_{9}y=\dfrac12\log_{3}y$, we get $\log_{3}(3^{1-x}+2)=1+\log_{3}(4\cdot 3^x-1)=\log_{3}!\big(3(4\cdot 3^x-1)\big)$. Hence $3^{1-x}+2=12\cdot 3^{x}-3$. Put $t=3^x$: $\dfrac{3}{t}+2=12t-3$. $\Rightarrow 12t^2-5t-3=0 \Rightarrow t=\dfrac{3}{4}$ (positive root). So $3^x=\dfrac{3}{4}\Rightarrow x=\log_{3}!\left(\dfrac{3}{4}\right)=1-\log_{3}4$.

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Solution

$00001101_2 = 13_{10}$ $00001111_2 = 15_{10}$ Product in decimal $= 13 \times 15 = 195$ Now convert $195_{10}$ to binary: $195 = 128 + 64 + 3 = 11000011_2$

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Solution

We use modulo properties: $27 \equiv 3 \pmod{12}$ $\Rightarrow 27^{40} \equiv 3^{40} \pmod{12}$ Now, $3^1 \equiv 3$, $3^2 \equiv 9$, $3^3 \equiv 3$, $3^4 \equiv 9$, pattern repeats every 2 powers. So, for even power $40$, remainder = $9$.