Educational

Proof of a^(q-r) b^(r-p) c^(p-q) = 1 Given \(a=xy^{p-1},\; b=xy^{q-1},\; c=xy^{r-1}\), prove: \[ a^{q-r} b^{r-p} c^{p-q} = 1 \] Proof \[ = (xy^{p-1})^{q-r} (xy^{q-1})^{r-p} (xy^{r-1})^{p-q} \] \[ = x^{q-r}y^{(p-1)(q-r)} \cdot x^{r-p}y^{(q-1)(r-p)} \cdot x^{p-q}y^{(r-1)(p-q)} \] \[ = x^{(q-r)+(r-p)+(p-q)} \cdot y^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)} \] \[ = x^0 \cdot y^0 \] \[ = 1 \] Hence Proved Next Question /

Simplify (16^-1/5)^(5/2) Simplify: \(\left(16^{-1/5}\right)^{5/2}\) Solution \[ = 16^{(-1/5)\cdot(5/2)} \] \[ = 16^{-1/2} \] \[ = \frac{1}{\sqrt{16}} = \frac{1}{4} \] Final Answer: \[ \boxed{\frac{1}{4}} \] Next Question / Full Exercise

Find a, b, c in 4725 = 3^a 5^b 7^c Given \(4725 = 3^a 5^b 7^c\), find: (i) \(a, b, c\) (ii) \(2^{-a} 3^b 7^c\) Solution \[ 4725 = 3^3 \times 5^2 \times 7^1 \] \[ \Rightarrow a = 3,\quad b = 2,\quad c = 1 \] (ii) Evaluate \[ 2^{-a} 3^b 7^c = 2^{-3}

Find a, b, c in 1176 = 2^a × 3^b × 7^c Find \(a, b, c\) if \(1176 = 2^a \times 3^b \times 7^c\) Solution \[ 1176 = 2^3 \times 3^1 \times 7^2 \] \[ \text{Comparing with } 2^a \times 3^b \times 7^c \] \[ a = 3,\quad b = 1,\quad c = 2 \]

Find a, b, c in 49392 = a^4 b^2 c^3 Find \(a, b, c\) if \(49392 = a^4 b^2 c^3\) Solution \[ 49392 = 2^5 \times 3^2 \times 7^3 \] \[ = (2)^5 (3)^2 (7)^3 \] \[ \text{Compare with } a^4 b^2 c^3 \] \[ a = 2,\quad b = 3,\quad c = 7 \]

Solve 4^(x-1)(0.5)^(3-2x) = (1/8)^x Solve: \(4^{x-1}\times (0.5)^{3-2x} = \left(\frac{1}{8}\right)^x\) Solution \[ 4 = 2^2,\quad 0.5 = 2^{-1},\quad \frac{1}{8} = 2^{-3} \] \[ = (2^2)^{x-1} \cdot (2^{-1})^{3-2x} = (2^{-3})^x \] \[ = 2^{2x-2} \cdot 2^{-3+2x} = 2^{-3x} \] \[ = 2^{4x-5} = 2^{-3x} \] \[ \Rightarrow 4x – 5 = -3x \] \[ \Rightarrow 7x =

Solve 3^(2x+4) + 1 = 2·3^(x+2) Solve: \(3^{2x+4} + 1 = 2\cdot3^{x+2}\) Solution \[ 3^{2x+4} + 1 = 2\cdot3^{x+2} \] \[ \text{Let } t = 3^{x+2} \] \[ \Rightarrow t^2 + 1 = 2t \] \[ \Rightarrow t^2 – 2t + 1 = 0 \] \[ \Rightarrow (t – 1)^2 = 0 \] \[ \Rightarrow

Solve 2^(2x) – 2^(x+3) + 2^4 = 0 Solve: \(2^{2x} – 2^{x+3} + 2^4 = 0\) Solution \[ 2^{2x} – 2^{x+3} + 16 = 0 \] \[ \text{Let } 2^x = t \] \[ \Rightarrow t^2 – 8t + 16 = 0 \] \[ \Rightarrow (t – 4)^2 = 0 \] \[ \Rightarrow t =

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