solve the simultaneous equations;
log x base 2-log y base 4=4,
log (x-2y)base 2=5
The second equation can be rewritten as log(x-2y) = 2^5
The first equation is more complicated because you have different bases. You need to chane base 4 to base 2 by using the rule that allows for base changes.
Once you do this, you can use the Law of Logs to combine the left side which will equal 2^4 or 16.
Can you finish it now? Do you need more help?
I have a strong feeling that Jessica has a typo , and all the bases are base 2
(assume when I write log (..) , I mean log2 (..) )
from the 1st:
log x - log y = 4 ----> x/y = 16 or x = 16y
from the 2nd:
log(x-2y) = 5 ---> x-2y =32
use substitution:
16 - 2y = 32
14y = 32
y = 16/7 , then x = 16(16/7) = 256/7
(256/7 , 16/7) satisfies both "assumed" equations)
To solve the simultaneous equations:
Equation 1: log(x) base 2 - log(y) base 4 = 4
Equation 2: log(x - 2y) base 2 = 5
Step 1: Convert equation 1 to a common base.
Using the logarithmic identity log(a) - log(b) = log(a/b), we can rewrite equation 1 as:
log(x/y) base 2 = 4
Step 2: Rewrite equation 2 in exponential form.
In exponential form, equation 2 becomes:
2^5 = x - 2y
32 = x - 2y → (Equation 3)
Step 3: Solve equation 3 for x - 2y and substitute it into equation 1.
Rearrange equation 3 to solve for x - 2y:
x - 2y = 32
Substitute the value of x - 2y into equation 1:
log(x/y) base 2 = 4 → log(32) base 2 = 4
Step 4: Solve the logarithmic equation.
Since log(32) base 2 = 5, we have:
x - 2y = 32 → (Equation 4)
Step 5: Solve equation 4 for x.
Rearrange equation 4 to solve for x:
x = 32 + 2y
Step 6: Substitute the value of x into equation 2.
Substitute the value of x into equation 2:
log(32 + 2y) base 2 = 5
Step 7: Solve the logarithmic equation.
Using the exponential form of logarithm, we have:
2^5 = 32 + 2y
32 = 32 + 2y
2y = 0
y = 0
Step 8: Substitute the value of y into equation 5.
Substitute the value of y into equation 5:
x = 32 + 2(0)
x = 32
Step 9: Check the solution.
Substitute the values of x and y into both original equations to check if they satisfy the equations.
Using equation 1: log(32) base 2 - log(0) base 4 = 4
log(32) base 2 - log(0) base 4 is undefined, as log(0) is undefined. Therefore, this solution is not valid.
The simultaneous equations have no valid solution.
To solve the given simultaneous equations:
First, let's rewrite the equations in exponential form using the property of logarithms:
1) log x base 2 - log y base 4 = 4
This can be rewritten as:
2^(log x base 2) - 4^(log y base 4) = 4
x - y^2 = 4
2) log (x-2y) base 2 = 5
This can be rewritten as:
2^(log (x-2y) base 2) = 2^5
x - 2y = 32
Now, we have a system of two equations:
x - y^2 = 4 ---(Equation 1)
x - 2y = 32 ---(Equation 2)
To solve this system, we can use a method called substitution:
From Equation 2, we can express x in terms of y:
x = 2y + 32
Substitute this value of x into Equation 1:
(2y + 32) - y^2 = 4
Rearrange the equation to form a quadratic equation:
y^2 - 2y - 28 = 0
Now, we can solve this quadratic equation using factoring or the quadratic formula. Once we find the values of y, we can substitute them back into Equation 2 to find the corresponding values of x.
I hope this explanation helps you understand how to solve the simultaneous equations involving logarithms.