Question 1135939: 216^-2k
-------- = 36 Please show steps and solution.
36^-2k-3 Found 4 solutions by Alan3354, Theo, MathTherapy, greenestamps:Answer by Alan3354(69443) (Show Source):
in step 2 i multiplied both sides of the equation by 36^(-2k-3)
in step 3 i converted 36^(-2k-3) to 36^(-2k) / 36^3 because they're equivalent.
in step 4 i divided 36 in the numerator into 36^3 in the denominator to get 36^2 in the denominator.
in step 5 i replaced (6 * 36)^(-2k) with 6^(-2k) * 36^(-2k) because they're equivalent.
in step 6 i divided both sides of the equation by 36^(-2k) which then removed 36^(-2k) from the equation because 36^(-2k) / 36^(-2k) = 1.
in step 7 i made 6^(-2k) equal to 1 / 6^(2k) because they're equivalent.
in step 8 i cross multiplied.
in step 9 i took the log of both sides of the equation to get log(36^2)( = log(6^(2k)) which then became 2 * log(36) = 2k * log(6) because log(x^a) = a*log(x).
in step 10 i divided both sides of the equation by log(6).
in step 11 i divided both sides of the equation by 2.
in step 12 i solved for k.
any questions how or why i got what i got, send email to dtheophilis@gmail.com.
in donfirmed by replacing k with 2 in the original equation and evaluated it to get 36 = 36.
If , then: ----- Applying
- 6k - (- 4k - 6) = 2 ------- Bases are equal and so are the exponents
- 6k + 4k + 6 = 2
- 2k = 2 - 6
- 2k = - 4
You DON'T NEED LOGS or any complex calculations....just the laws of exponents!
Using logs is just a waste of time and effort, unless you're requested to use logs, of course.
