1. Problem: Solve the exponential equation $2^{x-4} = 4a^{x-6}$ where $a > 0$ and $a \neq 1$. 2. Formula and rules: Recall that $4 = 2^2$, so rewrite the equation as $2^{x-4} = 2^2 \cdot a^{x-6}$. Since bases on the left are powers of 2, express the right side similarly if possible. 3. Intermediate steps: Rewrite the equation: $$2^{x-4} = 2^2 \cdot a^{x-6}$$ Divide both sides by $2^2$: $$2^{x-6} = a^{x-6}$$ If $x \neq 6$, take the $(x-6)$th root: $$2 = a$$ If $x=6$, check if the original equation holds: $$2^{6-4} = 2^2 = 4$$ Right side: $$4 \cdot a^{6-6} = 4 \cdot a^0 = 4$$ So $x=6$ is a solution for any $a$. 4. Final answer: The solution is $a=2$ and $x$ can be any real number, but specifically $x=6$ always satisfies the equation. --- 5. Problem: Expand $A = (1+2x)^7$ up to the fourth term and approximate $(0.99)^8$ to four decimal places. 6. Formula: Binomial expansion: $$(1 + y)^n = \sum_{k=0}^n \binom{n}{k} y^k$$ 7. Intermediate work: For $A = (1+2x)^7$, the first four terms are: $$\binom{7}{0}(2x)^0 = 1$$ $$\binom{7}{1}(2x)^1 = 7 \cdot 2x = 14x$$ $$\binom{7}{2}(2x)^2 = 21 \cdot 4x^2 = 84x^2$$ $$\binom{7}{3}(2x)^3 = 35 \cdot 8x^3 = 280x^3$$ For $(0.99)^8$, use binomial expansion with $x = -0.01$: $$(1 - 0.01)^8 = \sum_{k=0}^8 \binom{8}{k} (1)^{8-k} (-0.01)^k$$ Calculate first five terms for accuracy: $$1 - 8(0.01) + 28(0.01)^2 - 56(0.01)^3 + 70(0.01)^4$$ $$= 1 - 0.08 + 0.0028 - 0.000056 + 0.000007$$ Sum: $$= 0.922751$$ Rounded to four decimals: $0.9228$ --- 8. Problem: Find $x$ if the third term of the expansion of $(1+x)^8$ is $0.07$. 9. Formula: The $k$th term in $(1+x)^n$ is: $$T_{k+1} = \binom{n}{k} x^k$$ Third term means $k=2$: $$T_3 = \binom{8}{2} x^2 = 28 x^2$$ 10. Intermediate work: Set $28 x^2 = 0.07$ $$x^2 = \frac{0.07}{28} = 0.0025$$ $$x = \pm 0.05$$ 11. Final answer: $x = 0.05$ or $x = -0.05$