if 5^x 25^2y = 1 and 3^5x 9^y= 1/9,find the value of x and y.
5^x 25^2y = 1
5^x 5^4y = 1
5^(x+4y) = 1
x+4y = 0
3^5x 9^y = 1/9
3^5x 2^2y = 3^-2
5x+2y = -2
x = -4/9
y = 1/9
if (-2 1 x+2 )
(3 x-4 5 ) =25
(0 1 3 )
a) find the value of x
b) hence ,find (-2 1 x+2 )^-1
( 3 x-4 5 )
(o 1 3 )
p/s : all the value above IS COMBINE in one bracket ( )
YEAH,GO TO SPAIN STEAL EVERYTHING AND IDENTITY
Well, let's have some fun with these equations!
First, we can rewrite the equations as (5^x)(25^2y) = 1 and (3^5x)(9^y) = 1/9.
Now, let's simplify things a bit. We can write 25 as 5^2 and 9 as 3^2.
So the equations become (5^x)(5^4y) = 1 and (3^5x)(3^2y) = 1/9.
Using the rules of exponents, we can combine the numbers with the same base (5 and 3, respectively).
For the first equation, we get 5^(x+4y) = 1.
For the second equation, we get 3^(5x+2y) = 1/9.
Now, we know that any number raised to the power of 0 is equal to 1. So in both equations, we can set the exponents equal to 0.
For the first equation, we have x+4y = 0.
For the second equation, we have 5x+2y = 0.
Now, we can solve this system of equations. Let's double the first equation and subtract it from the second equation.
(2 * (x+4y)) - (5x+2y) = 0 - 0
2x + 8y - 5x - 2y = 0
-3x + 6y = 0
Now, we solve for x in terms of y: x = 2y/3.
Substituting this into the first equation, we get:
(2y/3) + (4y) = 0
Multiplying through by 3, we get:
2y + 12y = 0
14y = 0
So y = 0.
Now, substituting y back into x = 2y/3, we get x = 2(0)/3 = 0.
So the value of x is 0 and the value of y is also 0.
Hope that answers your question with a touch of clownish humor!
To find the values of x and y, we'll use the property of exponents that states:
a^m ⋅ b^n = (a ⋅ b)^(m+n)
Let's start by simplifying the given equations.
Given:
1) 5^x ⋅ 25^(2y) = 1
2) 3^(5x) ⋅ 9^y = 1/9
Let's simplify equation 1 first:
Since 25 = 5^2, we can rewrite equation 1 as:
5^x ⋅ (5^2)^(2y) = 1
Simplifying further, we get:
5^x ⋅ 5^(4y) = 1
Applying the exponent property, we add the exponents:
5^(x + 4y) = 1
Since any non-zero number raised to the power of 0 equals 1, we can conclude that:
x + 4y = 0 ----(Equation 3)
Now, let's simplify equation 2:
Since 9 = 3^2, we can rewrite equation 2 as:
3^(5x) ⋅ (3^2)^y = 1/9
Simplifying further, we get:
3^(5x) ⋅ 3^(2y) = 1/9
Applying the exponent property, we add the exponents:
3^(5x + 2y) = 1/9
Rewriting 1/9 as 3^(-2), we have:
3^(5x + 2y) = 3^(-2)
From the properties of exponents, we can conclude that:
5x + 2y = -2 ----(Equation 4)
Now we have a system of two equations (Equations 3 and 4) with two variables (x and y). We can solve this system simultaneously.
Taking Equation 3 and multiplying it by 2, we get:
2x + 8y = 0 ----(Equation 5)
Now, let's subtract Equation 5 from Equation 4:
(5x + 2y) - (2x + 8y) = -2 - 0
3x - 6y = -2
Simplifying,
3x - 6y = -2 ----(Equation 6)
Now we have two equations to solve simultaneously:
2x + 8y = 0 ----(Equation 5)
3x - 6y = -2 ----(Equation 6)
We can solve this system of linear equations using any preferred method, such as substitution or elimination.
Let's solve it using the elimination method:
Multiplying Equation 5 by 3 and Equation 6 by 2, we get:
6x + 24y = 0 ----(Equation 7)
6x - 12y = -4 ----(Equation 8)
Subtracting Equation 8 from Equation 7, we have:
(6x + 24y) - (6x - 12y) = 0 - (-4)
36y = 4
Dividing both sides of the equation by 4, we get:
y = 1/9
Now substituting the value of y in Equation 5, we can find the value of x:
2x + 8(1/9) = 0
2x + 8/9 = 0
Multiplying through by 9 to clear the fraction:
18x + 8 = 0
Subtracting 8 from both sides:
18x = -8
Dividing both sides by 18:
x = -8/18 = -4/9
Thus, the values of x and y are x = -4/9 and y = 1/9, respectively.