1. **State the problem:** Solve for $x$ in the equation $16(3)^2=81(2^x)$. 2. **Write down the equation:** $$16 \times 3^2 = 81 \times 2^x$$ 3. **Calculate $3^2$:** $$3^2 = 9$$ So the equation becomes: $$16 \times 9 = 81 \times 2^x$$ 4. **Multiply 16 and 9:** $$16 \times 9 = 144$$ So: $$144 = 81 \times 2^x$$ 5. **Isolate $2^x$ by dividing both sides by 81:** $$\frac{144}{81} = 2^x$$ Show cancellation: $$\frac{\cancel{144}}{\cancel{81}} = 2^x$$ Simplify fraction: $$\frac{16}{9} = 2^x$$ 6. **Rewrite $\frac{16}{9}$ as powers of 2 and 3:** $$\frac{16}{9} = \frac{2^4}{3^2}$$ 7. **Since the right side is $2^x$, and left side is $\frac{2^4}{3^2}$, the bases are different, so take logarithm base 2 of both sides:** $$x = \log_2\left(\frac{16}{9}\right) = \log_2(16) - \log_2(9)$$ 8. **Calculate each logarithm:** $$\log_2(16) = 4$$ $$\log_2(9) = \log_2(3^2) = 2 \log_2(3)$$ 9. **Final expression for $x$:** $$x = 4 - 2 \log_2(3)$$ **Answer:** $$x = 4 - 2 \log_2(3)$$