# MATHS - INDICES

by Shivangi
(Dubai)

Q) Solve for
x and y
3^(x+y) = 1/3

3^(x-y) = 1/√3

how do I solve this question ??

### Comments for MATHS - INDICES

 Dec 11, 2011 Indices Containing Variables by: Staff Question: by Shivangi (Dubai) Q) Solve for x and y 3^(x+y) = 1/3 3^(x-y) = 1/√3 how do I solve this question ?? Answer: Solving the system of equations: 1st equation: 3^(x + y) = 1/3 2nd equation: 3^(x - y) = 1/√3 ------------------------------------------------- Simplify the 1st equation: 3^(x + y) = 1/3 Multiply each side of the equation by 3 3 * 3^(x + y) = 3 * (1/3) 3 * 3^(x + y) = 1 * (3/3) 3 * 3^(x + y) = 1 * (1) 3 * 3^(x + y) = 1 3^(x + y + 1) = 1 Substitute 3^(0) in place of the 1 on the right side of the equation 3^(x + y + 1) = 3^(0) Take the log_base_3 of each side of the equation Log_base_3 [3^(x + y + 1)] = Log_base_3 [3^(0)] x + y + 1 = 0, this is the simplified form of the 1st equation ------------------------------- Simplify the 2nd equation: 3^(x - y) = 1/√3 Multiply each side of the equation by √3 √3 * [3^(x - y)] = √3 * (1/√3) √3 * [3^(x - y)] = 1 * (√3/√3) √3 * [3^(x - y)] = 1*(1) √3 * [3^(x - y)] = 1 (3^.5) * [3^(x - y)] = 1 3^(x - y + .5) = 1 Substitute 3^(0) in place of the 1 on the right side of the equation 3^(x - y + .5) = 3^(0) Take the log_base_3 of each side of the equation Log_base_3 [3^(x - y + .5)] = Log_base_3 [3^(0)] x - y + .5 = 0, this is the simplified form of the 2nd equation ----------------------------------- Solve the simplified system of equations 1st equation (simplified): x + y + 1 = 0 2nd equation (simplified): x - y + .5 = 0 Add the 1st equation and the 2nd equation, and then solve for x: x + y + 1 = 0 x - y + .5 = 0 ----------------- x + x + y - y + 1 + 0.5 = 0 + 0 (x + x) + (y – y) + (1 + 0.5) = (0 + 0) 2x + 0 + 1.5 = 0 2x + 1.5 = 0 Subtract 1.5 from each side of the equation 2x + 1.5 - 1.5 = 0 - 1.5 2x + 0 = -1.5 2x = -1.5 Divide each side of the equation by 2 2x/2 = -1.5/2 x * (2/2)= -1.5/2 x * (1)= -1.5/2 x = -1.5/2 x = -.75 = -3/4 Solve for y by substituting -.75 for x in the simplified form of the 2nd equation x - y + .5 = 0 -.75 - y + .5 = 0 Add y to each side of the equation -.75 - y + .5 + y = 0 + y -.75 - y + y + .5 = 0 + y -.75 + (- y + y) + .5 = 0 + y -.75 + 0 + .5 = 0 + y -.75 + .5 = 0 + y -.75 + .5 = 0 + y -.25 = 0 + y -.25 = y y = -.25 = -1/4 the FINAL ANSWER is: x = -3/4, y = -1/4 check the solution by substituting -.75 for x and -.25 for y in the original two equations: 1st equation: 3^(x + y) = 1/3 3^(x + y) = 1/3 3^(-.75 - .25) = 1/3 3^(-1) = 1/3 1/3 = 1/3, YES → the solutions for x and y are VALID SOLUTIONS are far as the 1st equation is concerned 2nd equation: 3^(x - y) = 1/√3 3^(x - y) = 1/√3 3^(-.75 - (-.25) = 1/√3 3^(-.75 + .25) = 1/√3 3^(-.5) = 1/√3 1/√3 = 1/√3 1/√3 = 1/√3, YES → the solutions for x and y are VALID SOLUTIONS are far as the 2nd equation is concerned Thanks for writing. Staff www.solving-math-problems.com