Educational

Solve system of exponential equations Solve: \[ 8^{x+1} = 16^{y+2}, \quad \left(\frac{1}{2}\right)^{3+x} = \left(\frac{1}{4}\right)^{3y} \] Solution \[ 8^{x+1} = 16^{y+2} \] \[ \Rightarrow (2^3)^{x+1} = (2^4)^{y+2} \] \[ \Rightarrow 2^{3x+3} = 2^{4y+8} \] \[ \Rightarrow 3x + 3 = 4y + 8 \quad …(1) \] \[ \left(\frac{1}{2}\right)^{3+x} = \left(\frac{1}{4}\right)^{3y} \] \[ \Rightarrow (2^{-1})^{3+x} = (2^{-2})^{3y} […]

Solve 3^(x-1) × 5^(2y-3) = 225 Solve: \(3^{x-1} \times 5^{2y-3} = 225\) Solution \[ 3^{x-1} \times 5^{2y-3} = 225 \] \[ \Rightarrow 3^{x-1} \times 5^{2y-3} = 3^2 \times 5^2 \] \[ \Rightarrow 3^{x-1} = 3^2,\quad 5^{2y-3} = 5^2 \] \[ \Rightarrow x-1 = 2 \] \[ \Rightarrow x = 3 \] \[ \Rightarrow 2y-3 =

Solve 4^(2x) = (cube root 16)^(-6/y) = (√8)^2 Solve: \(4^{2x} = (\sqrt[3]{16})^{-6/y} = (\sqrt{8})^2\) Solution \[ (\sqrt{8})^2 = (8^{1/2})^2 = 8 \] \[ \Rightarrow 4^{2x} = 8 \] \[ \Rightarrow (2^2)^{2x} = 2^3 \] \[ \Rightarrow 2^{4x} = 2^3 \] \[ \Rightarrow 4x = 3 \] \[ \Rightarrow x = \frac{3}{4} \] \[ (\sqrt[3]{16})^{-6/y} =

Solve 3^(x+1) = 27 × 3^4 Solve: \(3^{x+1} = 27 \times 3^4\) Solution \[ 3^{x+1} = 27 \times 3^4 \] \[ \Rightarrow 3^{x+1} = 3^3 \times 3^4 \] \[ \Rightarrow 3^{x+1} = 3^{7} \] \[ \Rightarrow x + 1 = 7 \] \[ \Rightarrow x = 6 \] Final Answer: \[ \boxed{x = 6} \]

Find x and y Find \(x\) and \(y\), if \(5^{3x} = 125\) and \(10^y = 0.001\) Solution \[ 5^{3x} = 125 \] \[ \Rightarrow 5^{3x} = 5^3 \] \[ \Rightarrow 3x = 3 \] \[ \Rightarrow x = 1 \] \[ 10^y = 0.001 \] \[ \Rightarrow 10^y = 10^{-3} \] \[ \Rightarrow y =

Find 2^(1+x) Find: \(2^{1+x}\), if \(3^{4x} = 81^{-1}\) and \(10^{1/y} = 0.0001\) Solution \[ 3^{4x} = 81^{-1} \] \[ \Rightarrow 3^{4x} = (3^4)^{-1} \] \[ \Rightarrow 3^{4x} = 3^{-4} \] \[ \Rightarrow 4x = -4 \] \[ \Rightarrow x = -1 \] \[ 10^{1/y} = 0.0001 \] \[ \Rightarrow 10^{1/y} = 10^{-4} \] \[ \Rightarrow

Find 2^(1+x) Find: \(2^{1+x}\), if \(3^{x+1} = 9^{x-2}\) Solution \[ 3^{x+1} = 9^{x-2} \] \[ \Rightarrow 3^{x+1} = (3^2)^{x-2} \] \[ \Rightarrow 3^{x+1} = 3^{2x-4} \] \[ \Rightarrow x+1 = 2x-4 \] \[ \Rightarrow x = 5 \] \[ 2^{1+x} = 2^{6} \] \[ = 64 \] Final Answer: \[ \boxed{64} \] Next Question /

Find (8x)^x Determine: \((8x)^x\), if \(9^{x+2} = 240 + 9^x\) Solution \[ 9^{x+2} = 240 + 9^x \] \[ \Rightarrow 9^x \cdot 9^2 = 240 + 9^x \] \[ \Rightarrow 81 \cdot 9^x = 240 + 9^x \] \[ \Rightarrow 81 \cdot 9^x – 9^x = 240 \] \[ \Rightarrow 80 \cdot 9^x = 240

Prove x^3 – 6x = 6 Given \(x = 2^{1/3} + 2^{2/3}\), show that \(x^3 – 6x = 6\) Solution \[ x = 2^{1/3} + 2^{2/3} \] \[ x^3 = (2^{1/3} + 2^{2/3})^3 \] \[ = (2^{1/3})^3 + (2^{2/3})^3 + 3 \cdot 2^{1/3} \cdot 2^{2/3}(2^{1/3} + 2^{2/3}) \] \[ = 2 + 4 + 3

Solve (√(3/5))^(x+1) = 125/27 Solve: \(\left(\sqrt{\frac{3}{5}}\right)^{x+1} = \frac{125}{27}\) Solution \[ \left(\sqrt{\frac{3}{5}}\right)^{x+1} = \frac{125}{27} \] \[ \Rightarrow \left(\frac{3}{5}\right)^{\frac{x+1}{2}} = \frac{5^3}{3^3} \] \[ \Rightarrow \frac{3^{\frac{x+1}{2}}}{5^{\frac{x+1}{2}}} = \frac{5^3}{3^3} \] \[ \Rightarrow 3^{\frac{x+1}{2}} \cdot 3^3 = 5^{\frac{x+1}{2}} \cdot 5^3 \] \[ \Rightarrow 3^{\frac{x+1}{2} + 3} = 5^{\frac{x+1}{2} + 3} \] \[ \Rightarrow \frac{x+1}{2} + 3 = 0 \] \[