1)If y varies inversely as x and y = 3, when x = 6. Find y when x = 36.
A)72
B)2
C)1/2
D)18
I chose C
y1/x2 = y2/x1
3/36 = y2/6
18/36 = 1/2
2)Write the equation 4^-3 = 1/64 in logarithmic form.
A)log64 = 4^3
B)log^-3 64 = 4
C)log^4 1/64 = -3
D)log^4(-3) = 64
I chose C
3)Solve: 6^n+1 >= 10. Round to four decimal places.
A)n >= 0.2851
B)n >= 0.6667
C)n >= 1.2851
D)n >= -0.7782
I chose C
log10/log6
1.2851
All answers are correct. Good work!
1) To solve this problem, we need to use the inverse variation equation which states that y varies inversely as x. The equation is usually written as y = k/x, where k is the constant of variation. Given that y = 3 when x = 6, we can find the value of k by substituting these values into the equation:
3 = k/6
To find k, we can cross multiply:
3 * 6 = k
k = 18
Now that we have the value of k, we can use it to find y when x = 36. Substitute the values into the equation:
y = 18/36
Simplifying, we get:
y = 1/2
So the answer is C) 1/2.
2) To write the equation 4^-3 = 1/64 in logarithmic form, we need to remember the property of logarithms that states that the logarithm of a number is the exponent to which another fixed value (the base) must be raised to produce that number.
In this case, the base is 4 and the number is 1/64. Using the logarithmic form of the equation, we get:
log4(1/64) = -3
So the answer is C) log^4 1/64 = -3.
3) To solve the inequality 6^n+1 >= 10, we first need to isolate the variable by subtracting 10 from both sides:
6^n+1 - 10 >= 0
Next, we can simplify the left side by using an exponent rule: a^(n+1) = a^n * a.
6^n * 6 - 10 >= 0
Simplifying further:
6^n * 6 - 10 >= 0
6^n * 6 >= 10
6^n >= 10/6
6^n >= 5/3
Now, we can take the logarithm (base 6) of both sides of the inequality to solve for n:
log6(6^n) >= log6(5/3)
Using the logarithmic property, we can bring down the exponent:
n >= log6(5/3)
Calculating this value using a calculator, we find:
n >= 0.2851
Rounding to four decimal places, the answer is A) n >= 0.2851.