Ravi Kant Kumar

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

Solve 13^(√x) = 4^4 – 3^4 – 6 Solve: \(13^{\sqrt{x}} = 4^4 – 3^4 – 6\) Solution \[ 13^{\sqrt{x}} = 4^4 – 3^4 – 6 \] \[ \Rightarrow 13^{\sqrt{x}} = 256 – 81 – 6 \] \[ \Rightarrow 13^{\sqrt{x}} = 169 \] \[ \Rightarrow 13^{\sqrt{x}} = 13^2 \] \[ \Rightarrow \sqrt{x} = 2 \] \[

Solve 5^(2x+3) = 1 Solve: \(5^{2x+3} = 1\) Solution \[ 5^{2x+3} = 1 \] \[ \Rightarrow 1 = 5^0 \] \[ \Rightarrow 5^{2x+3} = 5^0 \] \[ \Rightarrow 2x + 3 = 0 \] \[ \Rightarrow 2x = -3 \] \[ \Rightarrow x = -\frac{3}{2} \] Final Answer: \[ \boxed{x = -\frac{3}{2}} \] Next Question

Solve (cube root of 4)^(2x+1/2) = 1/32 Solve: \((\sqrt[3]{4})^{2x+\frac{1}{2}} = \frac{1}{32}\) Solution \[ (\sqrt[3]{4})^{2x+\frac{1}{2}} = \frac{1}{32} \] \[ \Rightarrow (4^{1/3})^{2x+\frac{1}{2}} = 2^{-5} \] \[ \Rightarrow 4^{\frac{2x+\frac{1}{2}}{3}} = 2^{-5} \] \[ \Rightarrow (2^2)^{\frac{2x+\frac{1}{2}}{3}} = 2^{-5} \] \[ \Rightarrow 2^{\frac{2(2x+\frac{1}{2})}{3}} = 2^{-5} \] \[ \Rightarrow 2^{\frac{4x+1}{3}} = 2^{-5} \] \[ \Rightarrow \frac{4x+1}{3} = -5 \] \[ \Rightarrow

Solve 2^(x-7) × 5^(x-4) = 1250 Solve: \(2^{x-7} \times 5^{x-4} = 1250\) Solution \[ 2^{x-7} \times 5^{x-4} = 1250 \] \[ \Rightarrow 2^{x-7} \times 5^{x-4} = 2 \times 5^4 \] \[ \Rightarrow \frac{2^x}{2^7} \times \frac{5^x}{5^4} = 2 \times 5^4 \] \[ \Rightarrow \frac{2^x \cdot 5^x}{2^7 \cdot 5^4} = 2 \cdot 5^4 \] \[ \Rightarrow 2^x

Solve 5^(x-2) × 3^(2x-3) = 135 Solve: \(5^{x-2} \times 3^{2x-3} = 135\) Solution \[ 5^{x-2} \times 3^{2x-3} = 135 \] \[ \Rightarrow 5^{x-2} \times 3^{2x-3} = 5 \times 3^3 \] \[ \Rightarrow \frac{5^x}{5^2} \times \frac{3^{2x}}{3^3} = 5 \times 3^3 \] \[ \Rightarrow \frac{5^x \cdot 3^{2x}}{5^2 \cdot 3^3} = 5 \cdot 3^3 \] \[ \Rightarrow 5^x

Solve (3/5)^x (5/3)^2x = 125/27 Solve: \(\left(\frac{3}{5}\right)^x \left(\frac{5}{3}\right)^{2x} = \frac{125}{27}\) Solution \[ \left(\frac{3}{5}\right)^x \left(\frac{5}{3}\right)^{2x} = \frac{125}{27} \] \[ \Rightarrow \frac{3^x}{5^x} \cdot \frac{5^{2x}}{3^{2x}} = \frac{125}{27} \] \[ \Rightarrow \frac{3^x \cdot 5^{2x}}{5^x \cdot 3^{2x}} = \frac{125}{27} \] \[ \Rightarrow \frac{5^x}{3^x} = \frac{125}{27} \] \[ \Rightarrow \left(\frac{5}{3}\right)^x = \left(\frac{5}{3}\right)^3 \] \[ \Rightarrow x = 3 \] Final Answer:

Solve (2^3)^4 = (2^2)^x Solve: \((2^3)^4 = (2^2)^x\) Solution \[ (2^3)^4 = (2^2)^x \] \[ = 2^{12} = 2^{2x} \] \[ \Rightarrow 12 = 2x \] \[ \Rightarrow x = 6 \] Final Answer: \[ \boxed{x = 6} \] Next Question / Full Exercise