What operation would you use in order to find x in this equation :3^x=243? And what about this : x^8=6561? If you wouldn't mind answering another question, why is n^0 always 1?
Just curious , that's all.
Just curious , that's all.
-
You must use a logarithm to solve 3^x = 243
xlog3 = log243
x = log243/log3
x = 5
To solve, x^8 = 6561, take the 8th root of 6561, which is more easily done by taking the square root of 6561 four times.
x^8 = 6561
x^4 = 81
x^2 = 9
x = (+ or -) 3
n^0 is always 1 because any number to the 0th power is 1.
xlog3 = log243
x = log243/log3
x = 5
To solve, x^8 = 6561, take the 8th root of 6561, which is more easily done by taking the square root of 6561 four times.
x^8 = 6561
x^4 = 81
x^2 = 9
x = (+ or -) 3
n^0 is always 1 because any number to the 0th power is 1.
-
you would use logarithims to solve the equation. so you would rewrite your equations as logs then solve. i cant answer the other question, i just learned that it was a rule that anything to the zero power is one.
-
3^x=3^5
x=5
or use logarithms (I find it easier to reduce to the same base when possible though)
6561^(1/8)
=3
because it acts like one.
example:
(3^3)/(3^3)=3^0
or
27/27=1
x=5
or use logarithms (I find it easier to reduce to the same base when possible though)
6561^(1/8)
=3
because it acts like one.
example:
(3^3)/(3^3)=3^0
or
27/27=1