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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 ??

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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



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