By first finding the value of \( 6^{x} \), solve the equation \( \frac{8^{1+1}}{9}=\frac{2^{2+1}}{3^{1}} \)

To solve the equation \[ \frac{8^{1+x}}{9} = \frac{2^{2x+1}}{3^{x}}, \] we can follow these steps, leveraging the value of \(6^{x}\) in the process. ### Step 1: Express All Terms with Prime Bases First, express all numbers in terms of their prime factors: - \(8 = 2^3\) - \(9 = 3^2\) Substituting these into the equation: \[ \frac{(2^3)^{1+x}}{3^2} = \frac{2^{2x+1}}{3^{x}} \] ### Step 2: Simplify the Exponents Apply the power rules (\((a^m)^n = a^{m \cdot n}\)): \[ \frac{2^{3(1+x)}}{3^2} = \frac{2^{2x+1}}{3^x} \] Simplify the exponents: \[ \frac{2^{3 + 3x}}{3^2} = \frac{2^{2x + 1}}{3^x} \] ### Step 3: Rearrange the Equation Multiply both sides by \(3^2 \cdot 3^x\) to eliminate the denominators: \[ 2^{3 + 3x} \cdot 3^x = 2^{2x + 1} \cdot 3^2 \] ### Step 4: Isolate the Terms Involving \(x\) Divide both sides by \(2^{2x + 1}\): \[ 2^{3 + 3x - (2x + 1)} \cdot 3^x = 3^2 \] Simplify the exponents: \[ 2^{2 + x} \cdot 3^x = 9 \] ### Step 5: Express in Terms of \(6^x\) Notice that \(6 = 2 \times 3\), so \(6^x = 2^x \cdot 3^x\). Rewrite the equation: \[ 2^{2} \cdot 2^x \cdot 3^x = 4 \cdot 6^x = 9 \] Thus: \[ 4 \cdot 6^x = 9 \] ### Step 6: Solve for \(6^x\) Divide both sides by 4: \[ 6^x = \frac{9}{4} \] ### Step 7: Solve for \(x\) To find \(x\), take the natural logarithm of both sides: \[ \ln(6^x) = \ln\left(\frac{9}{4}\right) \] Apply the logarithm power rule (\(\ln(a^b) = b \ln(a)\)): \[ x \ln(6) = \ln\left(\frac{9}{4}\right) \] Finally, solve for \(x\): \[ x = \frac{\ln\left(\frac{9}{4}\right)}{\ln(6)} \] ### Final Answer \[ x = \frac{\ln\left(\dfrac{9}{4}\right)}{\ln(6)} \] Or equivalently, \[ x = \log_6\left(\frac{9}{4}\right) \] ### Numerical Approximation (Optional) If you need a numerical value for \(x\): \[ x \approx \frac{\ln(2.25)}{\ln(6)} \approx \frac{0.81093}{1.79176} \approx 0.452 \]