Ravi Kant Kumar

Show that : (3^a/3^b)^{a+b} (3^b/3^c)^{b+c} (3^c/3^a)^{c+a} = 1

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

Show that : (3^a/3^b)^{a+b} (3^b/3^c)^{b+c} (3^c/3^a)^{c+a} = 1 Read More »

Simplify : [5^-1×7^2/5^2×7^-4]^7/2×[5^-2×7^3/5^3×7^-5]^-5/2

Simplify exponential expression Simplify: \[ \left[\frac{5^{-1}\cdot7^2}{5^2\cdot7^{-4}}\right]^{7/2} \times \left[\frac{5^{-2}\cdot7^3}{5^3\cdot7^{-5}}\right]^{-5/2} \] Solution \[ \left[\frac{5^{-1}}{5^2}\cdot\frac{7^2}{7^{-4}}\right]^{7/2} = \left[5^{-3}\cdot7^{6}\right]^{7/2} \] \[ = 5^{-21/2}\cdot7^{21} \] \[ \left[\frac{5^{-2}}{5^3}\cdot\frac{7^3}{7^{-5}}\right]^{-5/2} = \left[5^{-5}\cdot7^{8}\right]^{-5/2} \] \[ = 5^{25/2}\cdot7^{-20} \] \[ \text{Multiplying: } = 5^{-21/2+25/2}\cdot7^{21-20} \] \[ = 5^{2}\cdot7 \] \[ = 25\cdot7 = 175 \] Final Answer: \[ \boxed{175} \] Next Question / Full Exercise

Simplify : [5^-1×7^2/5^2×7^-4]^7/2×[5^-2×7^3/5^3×7^-5]^-5/2 Read More »