Here are the solutions to the exponential equations. 1. Solve for $x$: $(0.25)^{x+1} = 16$ Step 1: Rewrite $0.25$ and $16$ with a common base. $$0.25 = \frac{1}{4} = 4^{-1}$$ $$16 = 4^2$$ Step 2: Substitute these into the equation. $$(4^{-1})^{x+1} = 4^2$$ Step 3: Apply the exponent rule $(a^m)^n = a^{mn}$. $$4^{-(x+1)} = 4^2$$ $$4^{-x-1} = 4^2$$ Step 4: Equate the exponents since the bases are the same. $$-x - 1 = 2$$ Step 5: Solve for $x$. $$-x = 2 + 1$$ $$-x = 3$$ $$x = -3$$ The solution is $\boxed{-3}$. 2. Solve for $x$: $3(3)^x = 27$ Step 1: Isolate the exponential term by dividing both sides by $3$. $$3^x = \frac{27}{3}$$ $$3^x = 9$$ Step 2: Rewrite $9$ as a power of $3$. $$9 = 3^2$$ Step 3: Substitute this into the equation. $$3^x = 3^2$$ Step 4: Equate the exponents since the bases are the same. $$x = 2$$ The solution is $\boxed{2}$. 4. Solve for $x$: $10^{-x} = 0.000001$ Step 1: Rewrite $0.000001$ as a power of $10$. $$0.000001 = \frac{1}{1,000,000} = \frac{1}{10^6} = 10^{-6}$$ Step 2: Substitute this into the equation. $$10^{-x} = 10^{-6}$$ Step 3: Equate the exponents since the bases are the same. $$-x = -6$$ Step 4: Solve for $x$. $$x = 6$$ The solution is $\boxed{6}$. 5. Solve: $(3^x)^2 + 2(3^x) - 3 = 0$ Step 1: Let $y = 3^x$. Substitute $y$ into the equation. $$y^2 + 2y - 3 = 0$$ Step 2: Factor the quadratic equation. $$(y+3)(y-1) = 0$$ Step 3: Solve for $y$. $$y+3 = 0 \implies y = -3$$ $$y-1 = 0 \implies y = 1$$ Step 4: Substitute back $3^x$ for $y$ and solve for $x$. Case 1: $3^x = -3$ An exponential function with a positive base ($3$) can never be negative. Thus, there is no real solution for $x$ in this case. Case 2: $3^x = 1$ $$3^x = 3^0$$ Equate the exponents. $$x = 0$$ The solution is $\boxed{0}$. 6. Solve: $5^{2x+1} - 26(5^x) + 5 = 0$ Step 1: Rewrite $5^{2x+1}$ using exponent rules. $$5^{2x+1} = 5^{2x} \cdot 5^1 = (5^x)^2 \cdot 5$$ Step 2: Substitute this back into the equation. $$5(5^x)^2 - 26(5^x) + 5 = 0$$ Step 3: Let $y = 5^x$. Substitute $y$ into the equation. $$5y^2 - 26y + 5 = 0$$ Step 4: Factor the quadratic equation. $$5y^2 - 25y - y + 5 = 0$$ $$5y(y - 5) - 1(y - 5) = 0$$ $$(5y - 1)(y - 5) = 0$$ Step 5: Solve for $y$. $$5y - 1 = 0 \implies 5y = 1 \implies y = \frac{1}{5}$$ $$y - 5 = 0 \implies y = 5$$ Step 6: Substitute back $5^x$ for $y$ and solve for $x$. Case 1: $5^x = \frac{1}{5}$ $$5^x = 5^{-1}$$ Equate the exponents. $$x = -1$$ Case 2: $5^x = 5$ $$5^x = 5^1$$ Equate the exponents. $$x = 1$$ The solutions are $\boxed{-1, 1}$. Send me the next one 📸