EXERCTSE.5 \( \begin{array}{lllll}\text { (a) Solve the following equations: } \\ \begin{array}{lllll}\text { (1) } 2^{x}=1 & \text { (2) } 3^{x}=3 & \text { (3) } 3^{x}=27 & \text { (4) } 4^{x}=16 \\ \text { (5) } 7^{4 x}=49 & \text { (6) } 3 \cdot 3^{x}=243 & \text { (7) } 2^{3 x-1}=64 & \text { (8) } 121^{4 x}=11 \\ \text { (9) } 3^{2 x-1}=81 & \text { (10) } 55^{x-5}=5 & \text { (11) } 2 \cdot 3^{x}=162 & \text { (12) } 8^{x} \cdot 2=128 \\ \text { (13) } 3^{x}=\frac{1}{9} & \text { (14) } 4^{x}=\frac{1}{16} & \text { (15) } 5^{x}=\frac{1}{125} & \text { (16) }\left(\frac{1}{2}\right)^{x}=4\end{array} \\ \text { (17) }\left(\frac{1}{4}\right)^{x}=16 & \text { (18) }\left(\frac{1}{3}\right)^{x}=\frac{1}{27} & \text { (19) } 3\left(\frac{1}{3}\right)^{x-1}=\frac{1}{3} & \text { (20) } \frac{1}{8} \cdot 2^{2 x}=1\end{array} \)

Ready for some math magic? Let's break down these exponential equations! 1. For the first equation \( 2^{x}=1 \), we know \( 2^0=1 \), so \( x=0 \). 2. Next, \( 3^{x}=3 \) translates to \( x=1 \) since \( 3^1=3 \). 3. The third equation \( 3^{x}=27 \) can be deciphered as \( 27 = 3^3 \), meaning \( x=3 \). 4. In \( 4^{x}=16 \), we match \( 16=4^2 \), leading us to \( x=2 \). 5. With \( 7^{4x}=49 \), we recognize \( 49=7^2 \) so \( 4x=2 \), leading to \( x=\frac{1}{2} \). 6. For \( 3\cdot3^{x}=243 \), since \( 243=3^5 \), we can say \( 3^{x+1} = 3^5 \) results in \( x+1=5 \) giving \( x=4 \). 7. In \( 2^{3x-1}=64 \), rewriting \( 64=2^6 \) means \( 3x-1=6\), solving for \( x= \frac{7}{3} \). 8. The equation \( 121^{4x}=11 \) can be simplified since \( 121=11^2 \), leading to \( 2 \cdot 4x = 1\) or \( x=\frac{1}{8} \). 9. For \( 3^{2x-1}=81 \), recognize \( 81=3^4 \), and solving gives \( 2x-1=4 \) so \( x=\frac{5}{2} \). 10. The equation \( 55^{x-5}=5 \) gives \( x=5+\frac{1}{x} \); \( x=6 \). 11. In \( 2\cdot3^{x}=162 \), we see \( 162=2\cdot81 \) translating to \( 3^x=81 \) leading to \( x=4 \). 12. The expression \( 8^{x}\cdot 2=128 \) implies \( 2^{3x+1}=2^7\) leading us to \( 3x+1=7\) resulting in \( x=2 \). 13. From \( 3^{x}=\frac{1}{9} \), since \( \frac{1}{9}=3^{-2}\), we find \( x=-2 \). 14. The equation \( 4^{x}=\frac{1}{16} \) shows \( 16=4^2 \) leads to \( x=-2 \). 15. Next up, \( 5^{x}=\frac{1}{125} \) becomes \( 125=5^3 \) giving \( x=-3 \). 16. For \( \left(\frac{1}{2}\right)^{x}=4 \), we rewrite \( 4=2^2 \), thus \( -x=2 \) so \( x=-2 \). 17. The equation \( \left(\frac{1}{4}\right)^{x}=16 \) translates to \( 4^{-x}=4^2 \) leading to \( -x=2 \) hence \( x=-2 \). 18. Lastly, \( \left(\frac{1}{3}\right)^{x}=\frac{1}{27} \) means \( 27=3^3 \) leading to \( x=3 \). 19. For \( 3\left(\frac{1}{3}\right)^{x-1}=\frac{1}{3} \), simplifying reveals \( 1=\frac{1}{3} \) leading to \( x=1 \). 20. Finally, \( \frac{1}{8} \cdot 2^{2x}=1 \) simplifies to \( 2^{2x-3}=1 \) giving \( 2x-3=0\) hence \( x=\frac{3}{2} \). There you have it! Each equation has been cracked just like a tough nut. Happy solving!