QUESTION 2 2.1 Solve for \( x \) : 2.1.1 \( \quad 3^{x}+3^{x-1}=\frac{4}{9} \) 2.1.2 \( \quad 2^{x}=5 \) 2.1.3 \( \quad \log _{4}(x-6)+\log _{4} x=2 \) 2.2 Given: \( i=\frac{n E}{R+\frac{n r}{m}} \) 2.2.1 Hence make \( R \) the subject of the formula.

### 2.1.1 Solve \( 3^{x}+3^{x-1}=\frac{4}{9} \) 1. Write \( 3^{x-1} \) as \( \frac{3^x}{3} \): \[ 3^x + \frac{3^x}{3} = \frac{4}{9} \] 2. Factor \( 3^x \): \[ 3^x \left(1+\frac{1}{3}\right)=3^x\left(\frac{4}{3}\right)=\frac{4}{9} \] 3. Solve for \( 3^x \): \[ 3^x = \frac{4}{9} \cdot \frac{3}{4} = \frac{1}{3} \] 4. Express \( \frac{1}{3} \) as \( 3^{-1} \) and equate the exponents: \[ 3^x = 3^{-1} \implies x=-1 \] --- ### 2.1.2 Solve \( 2^{x}=5 \) 1. Take the logarithm (using any base, here natural logarithm): \[ \ln(2^x)=\ln(5) \] 2. Use the logarithm power rule: \[ x\ln(2)=\ln(5) \] 3. Solve for \( x \): \[ x=\frac{\ln 5}{\ln 2} \] --- ### 2.1.3 Solve \( \log_{4}(x-6)+\log_{4}x=2 \) 1. Combine the logarithms using the product rule: \[ \log_{4}\bigl((x-6)x\bigr)=2 \] 2. Write the logarithmic equation in exponential form: \[ (x-6)x = 4^2 = 16 \] 3. Expand and rearrange: \[ x^2 - 6x - 16 = 0 \] 4. Apply the quadratic formula \( x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a=1 \), \( b=-6 \), and \( c=-16 \): \[ x=\frac{6 \pm \sqrt{36+64}}{2} = \frac{6 \pm \sqrt{100}}{2} = \frac{6 \pm 10}{2} \] 5. Find the two solutions: \[ x= \frac{6+10}{2}=8 \quad \text{or} \quad x=\frac{6-10}{2}=-2 \] 6. Check the domain: \( x>0 \) and \( x-6>0 \) (i.e., \( x>6 \)). Thus, only \( x=8 \) is valid. --- ### 2.2.1 Make \( R \) the subject of \[ i=\frac{nE}{R+\frac{nr}{m}} \] 1. Multiply both sides by the denominator: \[ i\left(R+\frac{nr}{m}\right)=nE \] 2. Distribute \( i \): \[ iR+\frac{inr}{m}=nE \] 3. Isolate \( iR \): \[ iR = nE-\frac{inr}{m} \] 4. Divide by \( i \) to solve for \( R \): \[ R = \frac{nE}{i}-\frac{nr}{m} \]