solve the following equation for x:64 = 2 ^3x?

Answer 1

64 = 2 ^3x

that is 2^6 = 2 ^3x

therefore 6 = 3x

or x = 2

Answer 2

There are two ways to do this. One is using some tricks and the other is using logs.
Simple way:
64 is 2^ 6. So since the bases of both sides are the same, just worry about the powers that should eual each other-- 6 and 3x. 6 = 3x.
x = 2.
Now with logs:
log (64) to the base 2 = 3x
On a calculator, do log(64)/log(2)
That equals 6. So 6 = 3x.
Looks familiar? I think so!

Answer 3

x=2

Answer 4

First, observe that 2^6=64.

1. Then, the equation becomes 2^6=2^(3x) so 6=3x so x=2.

2. For an extra step, use log base 2. In the following, log means log base 2:

2^6=2^(3x)
log(2^6)=log(2^(3x))
6 log 2=3x log 2
6=3x (because log 2=1)
x=2

Note that the second method can be used for more general equations. So, if a=b^x, then ln(a)=ln(b^x)
ln(a)=x ln(b)
x=ln(a)/ln(b),

where here ln represents the natural logarithm function.

Answer 5

64 = 2^(3x)
2^6 = 2^(3x)
6 = 3x
x = 2

another way to think about this

64 = 2^(3x)
64 = (2^3)^x
64 = 8^x
8^2 = 8^x
x = 2

Answer 6

2

64 = 2 ^ 6 = 2 ^ (3x)

so

3x = 6, x = 2

Answer 7

64 = 2^3x
ln(64) = 3x*ln(2)
ln(64)/(3*ln(2)) = x = 2


Doug

Answer 8

x/64=2^3x

that's for sure.....

solve it to get x=2...hehehe