1. Solve the following simultaneous equations:
log2 xy = 7
log2 (x^2/y) = 5
2. If log y x =a and log z x=b where x is not equal to 1, express the following in terms of a and b:
logy (yz)
Solve the following simultaneous equations:
y = 2 log3 x
y+1 = log3 9x
Thanks! I tried to work them out but I couldn't!
Try this on the first set:
log (xy)= logx + logy
log (x^2y)= 2logx + log2
Those ought to help.
Try that also on the second.
On the third,
y+1= log3 9x = log33^2 x= 2 + log3 x
y-1= log3 x
but y= 2log3 x then solve.
I still cannot work them out!!
What problems are you having? Where are you getting stuck?
I don't know. I simply cannot agree with the answers and I have tried to work them out many times
I am having problems in mmeasuring.
1. To solve the simultaneous equations:
Equation 1: log2(xy) = 7
Equation 2: log2(x^2/y) = 5
Step 1: Rewrite the logarithmic equations in exponential form.
In general, if log base a of b = c, then a^c = b.
Using this, rewrite Equation 1 as: xy = 2^7 = 128
For Equation 2, the exponentiation might seem a bit trickier due to the division. However, we can use the fact that log2(x^2/y) = log2(x^2) - log2(y). So we have:
x^2 = y * 2^5 = 32y
Step 2: Substitute Equation 1 into Equation 2.
Replace xy with its value from Equation 1 in Equation 2:
x^2 = 32y
x * 128 = 32y
x = 32y/128
x = y/4
Step 3: Substituting x in terms of y into Equation 1, we get:
(xy)/4 = 128
y^2/4 = 128
y^2 = 512
y = ±√512
y = ±16√2
Step 4: Substitute the values of y back into Equation 1 to get the corresponding values of x:
When y = 16√2,
x = (16√2)/4 = 4√2
When y = -16√2,
x = (-16√2)/4 = -4√2
Thus, the solution to the simultaneous equations is (x, y) = (4√2, 16√2) or (-4√2, -16√2).
2. To express logy (yz) in terms of a and b, we can use the properties of logarithms:
log y (yz) = log y z + log y y
Since log y y = 1, we have:
log y (yz) = log y z + 1
Now, we can express log y z in terms of a and b using the given information:
log y z = log z z / log z y
From the given information, log y x = a and log z x = b. Therefore, log z y = 1 / log y z = 1 / a.
Substituting this back into the expression for log y (yz):
log y (yz) = log y z + 1
= 1/a + 1
So, logy (yz) can be expressed as (1/a + 1).
3. To solve the simultaneous equations:
Equation 1: y = 2 log3 x
Equation 2: y + 1 = log3 (9x)
Step 1: Apply the properties of logarithms to simplify the equations:
Equation 1: y = log3 x^2
Equation 2: y + 1 = log3 9x = log3 (3^2 * x) = log3 3^2 + log3 x = 2 + log3 x
Step 2: Set the right-hand sides of both equations equal to each other:
log3 x^2 = 2 + log3 x
Step 3: Rearrange the equation to get all terms on one side:
log3 x^2 - log3 x = 2
Step 4: Apply the logarithmic identity loga(b) - loga(c) = loga(b/c):
log3 (x^2 / x) = 2
Step 5: Simplify the expression inside the logarithm:
log3 x = 2
Step 6: Apply the inverse logarithmic function (3^2 = 9):
x = 9
Step 7: Substitute the value of x into Equation 2 to find y:
y + 1 = log3 (9x)
y + 1 = log3 (9 * 9)
y + 1 = log3 81
y + 1 = 4
Step 8: Solve for y:
y = 4 - 1
y = 3
The solution to the simultaneous equations is x = 9 and y = 3.