\(\begin{aligned} & \frac{1}{5^{-x}}=625 \\ \Rightarrow & 5^{x}=625 \\ \Rightarrow & 5^{x}=5^{4} \\ \Rightarrow & x=4 \end{aligned}\)

এখানে,

\(\begin{aligned} & \frac{1}{3^{-x}}=81 \\ \Rightarrow & 3^{x}=81 \\ \Rightarrow & 3^{x}=3^{4} \\ \Rightarrow & x=9\end{aligned}\)

\(\begin{aligned} & a^{n+1} \cdot a^{-n} \\ = & a^{n+1-n} \\ = & a^{1} \\ = & a\end{aligned}\)

এখানে,

\(\begin{aligned} 2^{x} & =\frac{1}{8} \\ \Rightarrow 2^{x} & =\frac{1}{2^{3}} \\ \Rightarrow 2^{x} & =2^{-3} \\ \therefore x & =-3 \end{aligned}\)

\(\text { } \begin{aligned} & (\sqrt{2})^{x-1}=16 \\ \Rightarrow & (2)^{\frac{x-1}{2}}=2^{4} \\ \Rightarrow & \frac{x-1}{2}=4 \\ \Rightarrow & x-1=8 \\ \Rightarrow & x=8+2=9 \end{aligned}\)

Ans

\(27 = 3^3\)

\(\sqrt[3]{27} = \sqrt[3]{3^3} = 3\)

\(\begin{array}{l}\left(2 x^{-1} \sqrt[3]{x^{2}}\right)^{-6}=\left(2 \cdot \frac{1}{x} \cdot x^{\frac{2}{3}}\right)^{-6} \\ =\left(2 \cdot x^{-\frac{1}{3}}\right)^{-6}=2^{-6} x^{2}=\frac{x^{2}}{64}\end{array}\)

\(\begin{aligned} & (\sqrt{3})^{x+1}=27 \\ \Rightarrow & 3^{\frac{x+1}{2}}=3^{3} \\ \Rightarrow & \frac{x+1}{2}=3 \\ \Rightarrow & x+1=6 \\ \Rightarrow & x=6-1=5\end{aligned}\)

দেওয়া আছে,

\(2.8^{x-1} = 1\)

আমরা জানি, যেকোনো অশুন্য সংখ্যার পাওয়ার বা ঘাত 0 হলে তার মান 1 হয়। অর্থাৎ, \(a^0 = 1\) (যেখানে \(a \neq 0\))।

সুতরাং, আমরা লিখতে পারি:

\(2.8^{x-1} = 2.8^0\)

যেহেতু উভয় পাশের ভিত্তি (base) সমান, তাই তাদের ঘাতও সমান হবে:

\(x-1=0\)

\(\therefore x = 1\)

প্রদত্ত রাশি,

\(\begin{array}{l}=\frac{7 \cdot 2^{x+1}-13 \cdot 2^{x}}{2^{x}} \\ =\frac{7 \cdot 2^{x} \cdot 2-13 \cdot 2^{x}}{2^{x}} \\ =\frac{2^{x}(7 \cdot 2-13)}{2^{x}} \\ =14-13 \\ =1\end{array}\)

এখানে,

\(\begin{aligned} & -16^{3 / 4} \\ = & \left(2^{4}\right)^{3 / 4} \\ = & 2^{\frac{3 \times 4}{4}} \\ = & 2^{3} \\ = & 8 \end{aligned}\)

\(\begin{aligned} & 8^{x+3}=64 \\ \Rightarrow & 8^{x+3}=8² \\ \Rightarrow & x+3=2 \\ \Rightarrow & x=2-3=-1 \\ & \therefore x=-1\end{aligned}\)