Ravi Kant Kumar

Simplify exponential expression Simplify: \[ \frac{6\cdot8^{n+1} + 16\cdot2^{3n-2}}{10\cdot2^{3n-1} – 7\cdot8^n} \] Solution \[ 8 = 2^3 \] \[ = \frac{6\cdot(2^3)^{n+1} + 16\cdot2^{3n-2}}{10\cdot2^{3n-1} – 7\cdot(2^3)^n} \] \[ = \frac{6\cdot2^{3n+3} + 2^4\cdot2^{3n-2}}{10\cdot2^{3n-1} – 7\cdot2^{3n}} \] \[ = \frac{2^{3n+2}(12 + 1)}{2^{3n-1}(10 – 14)} \] \[ = \frac{13\cdot2^{3n+2}}{-4\cdot2^{3n-1}} \] \[ = -13\cdot2^3 \] \[ = -104 \] Final Answer: […]

Simplify given exponential expression Simplify: \[ \frac{5\cdot25^{n+1}-25\cdot5^{2n}}{5\cdot5^{2n+3}-25^{n+1}} \] Solution \[ 25 = 5^2 \] \[ = \frac{5\cdot(5^2)^{n+1} – 5^2\cdot5^{2n}}{5\cdot5^{2n+3} – (5^2)^{n+1}} \] \[ = \frac{5^{2n+3} – 5^{2n+2}}{5^{2n+4} – 5^{2n+2}} \] \[ = \frac{5^{2n+2}(5 – 1)}{5^{2n+2}(25 – 1)} \] \[ = \frac{4}{24} = \frac{1}{6} \] Final Answer: \[ \boxed{\frac{1}{6}} \] Next Question / Full Exercise

Simplify (3^n × 9^(n+1)) / (3^(n-1) × 9^(n-1)) Simplify: \(\frac{3^n \times 9^{n+1}}{3^{n-1} \times 9^{n-1}}\) Solution \[ = \frac{3^n \cdot (3^2)^{n+1}}{3^{n-1} \cdot (3^2)^{n-1}} \] \[ = \frac{3^n \cdot 3^{2n+2}}{3^{n-1} \cdot 3^{2n-2}} \] \[ = 3^{\,n+2n+2-(n-1+2n-2)} \] \[ = 3^{\,3n+2-(3n-3)} = 3^5 \] \[ = 243 \] Final Answer: \[ \boxed{243} \] Next Question / Full Exercise

Proof of exponent identity Prove: \[ \left(\frac{x^a}{x^b}\right)^{a^2+ab+b^2} \times \left(\frac{x^b}{x^c}\right)^{b^2+bc+c^2} \times \left(\frac{x^c}{x^a}\right)^{c^2+ca+a^2} = 1 \] Proof \[ = x^{(a-b)(a^2+ab+b^2)} \cdot x^{(b-c)(b^2+bc+c^2)} \cdot x^{(c-a)(c^2+ca+a^2)} \] \[ = x^{(a^3-b^3) + (b^3-c^3) + (c^3-a^3)} \] \[ = x^{0} \] \[ = 1 \] Hence Proved Next Question / Full Exercise

Proof of (a^-1 + b^-1)^-1 = ab/(a+b) Prove: \((a^{-1}+b^{-1})^{-1} = \frac{ab}{a+b}\) Proof \[ (a^{-1}+b^{-1})^{-1} \] \[ = \left(\frac{1}{a} + \frac{1}{b}\right)^{-1} \] \[ = \left(\frac{a+b}{ab}\right)^{-1} \] \[ = \frac{ab}{a+b} \] Hence Proved Next Question / Full Exercise

Proof of (a+b+c)/(a^-1 + b^-1 + c^-1) = abc Prove: \(\frac{a+b+c}{a^{-1}+b^{-1}+c^{-1}} = abc\) Proof \[ = \frac{a+b+c}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \] \[ = \frac{a+b+c}{\frac{bc+ca+ab}{abc}} \] \[ = (a+b+c)\cdot \frac{abc}{ab+bc+ca} \] \[ = abc \] Hence Proved Next Question / Full Exercise

Solve 2^(3x-7) = 256 Solve: \(2^{3x-7} = 256\) Solution \[ 2^{3x-7} = 256 \] \[ 256 = 2^8 \] \[ \Rightarrow 2^{3x-7} = 2^8 \] \[ \Rightarrow 3x – 7 = 8 \] \[ \Rightarrow 3x = 15 \] \[ \Rightarrow x = 5 \] Final Answer: \[ \boxed{x = 5} \] Next Question /

Solve 4^(2x) = 1/32 Solve: \(4^{2x} = \frac{1}{32}\) Solution \[ 4^{2x} = \frac{1}{32} \] \[ (2^2)^{2x} = 2^{-5} \] \[ 2^{4x} = 2^{-5} \] \[ \Rightarrow 4x = -5 \] \[ \Rightarrow x = -\frac{5}{4} \] Final Answer: \[ \boxed{x = -\frac{5}{4}} \] Next Question / Full Exercise

Solve 2^(5x+3) = 8^(x+3) Solve: \(2^{5x+3} = 8^{x+3}\) Solution \[ 2^{5x+3} = (2^3)^{x+3} \] \[ = 2^{5x+3} = 2^{3x+9} \] \[ \Rightarrow 5x + 3 = 3x + 9 \] \[ \Rightarrow 2x = 6 \] \[ \Rightarrow x = 3 \] Final Answer: \[ \boxed{x = 3} \] Next Question / Full Exercise

Solve 2^(x+1) = 4^(x-3) Solve: \(2^{x+1} = 4^{x-3}\) Solution \[ 2^{x+1} = (2^2)^{x-3} \] \[ = 2^{x+1} = 2^{2x-6} \] \[ \Rightarrow x + 1 = 2x – 6 \] \[ \Rightarrow x = 7 \] Final Answer: \[ \boxed{x = 7} \] Next Question / Full Exercise