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

**(1) Solve \(2^x=1\)** - We know that any nonzero number raised to the 0 power is 1. - Thus, \(2^x=2^0\), so \[ x=0. \] **(2) Solve \(3^x=3\)** - Write \(3\) as \(3^1\). - Thus, \(3^x=3^1\), so \[ x=1. \] **(3) Solve \(3^x=27\)** - Since \(27=3^3\), we have \[ 3^x=3^3, \] which implies \[ x=3. \] **(4) Solve \(4^x=16\)** - Notice that \(16=4^2\). - So, \(4^x=4^2\) and hence \[ x=2. \] **(5) Solve \(7^{4x}=49\)** - Write \(49\) as \(7^2\). - Then, \(7^{4x}=7^2\) gives \[ 4x=2,\quad\text{so}\quad x=\frac{1}{2}. \] **(6) Solve \(3\cdot3^x=243\)** - Write \(243\) as \(3^5\). - Notice that \(3\cdot3^x=3^{x+1}\), so \[ 3^{x+1}=3^5. \] - Therefore, \[ x+1=5\quad\text{and}\quad x=4. \] **(7) Solve \(2^{3x-1}=64\)** - Write \(64\) as \(2^6\). - Thus, \(2^{3x-1}=2^6\) leads to \[ 3x-1=6\quad\Rightarrow\quad 3x=7,\quad x=\frac{7}{3}. \] **(8) Solve \(121^{4x}=11\)** - Express \(121\) as \(11^2\); then: \[ (11^2)^{4x}=11\quad\Rightarrow\quad 11^{8x}=11^1. \] - Equating the exponents: \[ 8x=1\quad\Rightarrow\quad x=\frac{1}{8}. \] **(9) Solve \(3^{2x-1}=81\)** - Write \(81\) as \(3^4\). - Thus, \(3^{2x-1}=3^4\) implies \[ 2x-1=4\quad\Rightarrow\quad 2x=5,\quad x=\frac{5}{2}. \] **(10) Solve \(55^{x-5}=5\)** - Taking the natural logarithm on both sides: \[ \ln\left(55^{x-5}\right)=\ln5. \] - This gives \[ (x-5)\ln55=\ln5, \] so \[ x-5=\frac{\ln5}{\ln55}\quad\Rightarrow\quad x=5+\frac{\ln5}{\ln55}. \] **(11) Solve \(2\cdot3^x=162\)** - Divide both sides by 2: \[ 3^x=\frac{162}{2}=81. \] - Since \(81=3^4\), we have \[ 3^x=3^4,\quad\text{hence}\quad x=4. \] **(12) Solve \(8^x\cdot2=128\)** - Divide both sides by 2: \[ 8^x=\frac{128}{2}=64. \] - Recognize that \(64=8^2\), so \[ 8^x=8^2,\quad\text{and}\quad x=2. \] **(13) Solve \(3^x=\frac{1}{9}\)** - Write \(\frac{1}{9}\) as \(3^{-2}\). Thus: \[ 3^x=3^{-2}\quad\Rightarrow\quad x=-2. \] **(14) Solve \(4^x=\frac{1}{16}\)** - Since \(16=4^2\), \(\frac{1}{16}=4^{-2}\). So, \[ 4^x=4^{-2}\quad\Rightarrow\quad x=-2. \] **(15) Solve \(5^x=\frac{1}{125}\)** - Recognize that \(125=5^3\), so \(\frac{1}{125}=5^{-3}\). Therefore, \[ 5^x=5^{-3}\quad\Rightarrow\quad x=-3. \] **(16) Solve \(\left(\frac{1}{2}\right)^x=4\)** - Express the left side as \(2^{-x}\) and note that \(4=2^2\). Thus, \[ 2^{-x}=2^2. \] - Equate the exponents: \[ -x=2,\quad\text{so}\quad x=-2. \] **(17) Solve \(\left(\frac{1}{4}\right)^x=16\)** - Write \(\left(\frac{1}{4}\right)^x\) as \(4^{-x}\) and recognize \(16=4^2\). So, \[ 4^{-x}=4^2. \] - Equate exponents: \[ -x=2,\quad\text{hence}\quad x=-2. \] **(18) Solve \(\left(\frac{1}{3}\right)^x=\frac{1}{27}\)** - Express \(\left(\frac{1}{3}\right)^x\) as \(3^{-x}\) and write \(\frac{1}{27}\) as \(3^{-3}\). Thus: \[ 3^{-x}=3^{-3}. \] - Equate the exponents: \[ -x=-3,\quad\text{so}\quad x=3. \] **(19) Solve \(3\left(\frac{1}{3}\right)^{x-1}=\frac{1}{3}\)** - Write \(\left(\frac{1}{3}\right)^{x-1}\) as \(3^{-(x-1)}\). Then the equation becomes: \[ 3\cdot3^{-(x-1)}=3^{1-(x-1)}=3^{2-x}. \] - Since \(\frac{1}{3}=3^{-1}\), we have: \[ 3^{2-x}=3^{-1}. \] - Equate the exponents: \[ 2-x=-1,\quad\text{so}\quad x=3. \] **(20) Solve \(\frac{1}{8}\cdot2^{2x}=1\)** - Multiply both sides by 8: \[ 2^{2x}=8. \] - Since \(8=2^3\), equate the exponents: \[ 2x=3,\quad\text{so}\quad x=\frac{3}{2}. \]