Solve simultaneous equation.
log2(x-14y)=3
logx-log(y+1)=1
I noticed a typo in the above solution.
y+10 -14y=8 should be
10y + 10 - 14y = 8
The solution comes out a bit "nicer".
To solve the simultaneous equations:
Equation 1: log2(x-14y)=3
Equation 2: logx-log(y+1)=1
Step 1: Rewrite the logarithmic equations in exponential form.
In exponential form, Equation 1 becomes: 2^3 = (x-14y) -> 8 = x - 14y
In exponential form, Equation 2 becomes: x = (y+1)^1 -> x = (y+1)
Step 2: Substitute the value of x from Equation 2 into Equation 1.
Substituting x = (y+1), Equation 1 becomes: 8 = (y+1) - 14y
Simplify the equation: 8 = 1 - 13y
Step 3: Solve for y.
Rearrange the equation: 13y = 1 - 8
Simplify: 13y = -7
Divide both sides by 13: y = -7/13
Step 4: Substitute the value of y into Equation 2 to find x.
Substituting y = -7/13, Equation 2 becomes: x = (-7/13 + 1)
Simplify: x = (6/13)
Step 5: Final solution
The simultaneous equations are satisfied when x = 6/13 and y = -7/13.
To solve the given system of simultaneous equations:
Equation 1: log2(x-14y) = 3
Equation 2: logx - log(y+1) = 1
Let's solve equation 1 first.
1. Apply the properties of logarithms to rewrite equation 1:
2^3 = x - 14y (since log base 2 of a number is equivalent to writing the number as 2 raised to the power of the logarithm)
8 = x - 14y
Now let's solve equation 2.
2. Use the quotient property of logarithms to rewrite equation 2:
log(x / (y+1)) = 1
3. Rewrite the logarithm equation in exponential form:
10^1 = x / (y+1)
10 = x / (y+1)
Now we have two equations:
Equation 1: x - 14y = 8
Equation 2: x = 10(y+1)
We can now solve the system of equations using substitution or elimination method.
Let's use substitution:
Substitute the value of x from equation 2 into equation 1:
10(y+1) - 14y = 8
10y + 10 - 14y = 8
-4y + 10 = 8
-4y = 8 - 10
-4y = -2
y = (-2 / -4)
y = 0.5
Now substitute the value of y back into equation 2 to find x:
x = 10(0.5 + 1)
x = 10(1.5)
x = 15
Therefore, the solution to the system of simultaneous equations is x = 15 and y = 0.5.
The first equation is easy, if log base 2 is 3, then
x-14y=8
The second equation is the same as
x/(y+1)= 10 or x =10y+10
then
y+10 -14y=8
-13y=-2
and you have y
then x=10(2/13)+10=11.53
y=.154