Find the values of x and y in the two equations below then answer the question that follows.
2^-3=1/x
7^y=1/49
What is the value of xy ?
For the first problem i got 8.
For the second one i got 1/343
I don't know what im doing wrong or right. Can you please show me with work.
ive got it ive seen my mistake
Answer is -16 :)
2^-3=1/x
2^3 = x
x = 8, you are correct
7^y=1/49
7^y = 1/7^2
t^y = 7^-2,
thus y = -2
To find the values of x and y in the given equations, we can solve them step by step.
First equation: 2^-3 = 1/x
To find x, we need to isolate it on one side of the equation.
We know that 2^-3 is equal to (1/2)^3, which is 1/8. We substitute that in:
1/8 = 1/x
To solve for x, we can take the reciprocal of both sides:
8/1 = x
Hence, x = 8.
Second equation: 7^y = 1/49
We want y, so we take the logarithm of both sides. Let's use the base-10 logarithm:
log(7^y) = log(1/49)
Using the logarithmic property, we can bring down the exponent:
y * log(7) = log(1/49)
Next, we divide both sides of the equation by log(7):
y = log(1/49) / log(7)
Now, let's calculate this value:
y ≈ -2
Therefore, x = 8 and y ≈ -2.
Now we can find the value of xy:
xy = 8 * (-2) = -16
So, the value of xy is -16.