solve for x and y , the simultaneous equations 4^x+3 = 32(2^x+y) and 9^x + 3^y = 10
Assuming the usual carelessness with parentheses, I get
4^(x+3) = 32(2^(x+y))
2^(2(x+3)) = 2^5*2^(x+y)
2x+6 = 5+x+y
x = y-1
9^(y-1) + 3^y = 10
3^(2y-2) + 3^y = 10
3^(2y)/9 + 3^y = 10
3^(2y) + 9*3^y - 90 = 0
(3^y + 15)(3^y - 6) = 0
3^y = 6
y = log6/log3
x = y-1
That answer is unusual, so check to make sure I got things right.
Right answer. The answer is:
y = 1.6309, x = 0.6309
How did u get (3^y + 15)(3^y - 6) = 0 ?
I didn't understand how did you get (3^y+15) (3^y-6)
9^x + 3^y =10. [x=y-1]
3^2x + 3^y =10
3^(2y-2) + 3^y =10
3^2y / 3^2 + 3^y = 10
3^2y + (3^2)(3^y) =10
Here assume 3^y=u
u² + 9u -10=0
Using quadratic function
u=6 or -15
(Skip -ve value )
u=6. 3^y=6
log base 3 of 6= y
Change base value
lg6/lg3 =y
y=1.63
x=0.63
To solve the simultaneous equations 4^x+3 = 32(2^x+y) and 9^x + 3^y = 10, we can follow these steps:
Step 1: Simplify the equations
Let's simplify the expressions in both equations.
In the first equation, simplify 4^x+3:
4^x+3 = 32(2^x+y)
2^(2x) * 2^3 = 32(2^x+y)
2^(2x + 3) = 32 · 2^x · 2^y
2^(2x + 3) = 2^(5 + x + y)
By comparing the exponents, we get:
2x + 3 = 5 + x + y ...(Equation 1)
In the second equation, simplify 9^x:
9^x + 3^y = 10
(3^2)^x + 3^y = 10
3^(2x) + 3^y = 10
Step 2: Substitute the simplified expressions
Since equation 1 gives us a relationship between x, y, and x, let's substitute (x + y = 2x + 3 - 5) into equation 2:
3^(2x) + 3^y = 10
3^(2x) + 3^(2x + 3 - 5) = 10 (Using x + y = 2x + 3 - 5)
Simplifying further, we get:
3^(2x) + 3^(2x - 2) = 10
We now have a single equation in terms of x.
Step 3: Solve for x
Let's solve the equation 3^(2x) + 3^(2x - 2) = 10 for x.
By rearranging the equation:
3^(2x) + 3^(2x) * 3^(-2) = 10
3^(2x)(1 + 3^(-2)) = 10
3^(2x)(1 + 1/9) = 10
3^(2x)(10/9) = 10
3^(2x) = 10 * 9/10
3^(2x) = 9
To solve for x, we take the logarithm (base 3) on both sides:
2x = log₃(9)
x = log₃(9) / 2
Step 4: Solve for y
Now that we have found the value of x, we can substitute it back into equation 1 to solve for y.
Using equation 1: 2x + 3 = 5 + x + y
Substituting for x: 2 * (log₃(9) / 2) + 3 = 5 + log₃(9) / 2 + y
Simplifying:
log₃(9) + 3 = 5 + log₃(9) / 2 + y
3 - 5 = y + log₃(9) / 2 - log₃(9)
-2 = y - log₃(9) / 2
Combining the terms:
-2 = 2y - log₃(9) / 2
Solving for y:
2y = -2 + log₃(9) / 2
y = (-2 + log₃(9) / 2) / 2
Therefore, the values of x and y that satisfy the simultaneous equations are:
x = log₃(9) / 2
y = (-2 + log₃(9) / 2) / 2