It is helpful to realize that
Log(baseX)[Y] = Log[Y]/Log[X] (where the logs on the right are both to any same base)
and
Log(baseB)[B^A] = A (from the definition of a logarithm)
Your expression becomes
x = Log[1/2]/Log[16]*Log[125]/Log[25]*Log[81]/Log[27]
You can use your calculator to work this out, or you can make use of additional facts you know. These are
1/2 = 2^(-1); 16 = 2^4; so Log[1/2]/Log[16] = -1/4 when the base of each log is 2.
125 = 5^3; 25 = 5^2; so Log[125]/Log[25] = 3/2 when the base of each log is 5.
81 = 3^4; 27 = 3^3; so Log[81]/Log[27] = 4/3 when the base of each log is 3.
Now we can find x easily as the product of these fractions.
x = (-1/4)*(3/2)*(4/3) = (-1*3*4)/(4*2*3) = (-1/2)*(3*4)/(3*4)
x = -1/2
Log(baseX)[Y] = Log[Y]/Log[X] (where the logs on the right are both to any same base)
and
Log(baseB)[B^A] = A (from the definition of a logarithm)
Your expression becomes
x = Log[1/2]/Log[16]*Log[125]/Log[25]*Log[81]/Log[27]
You can use your calculator to work this out, or you can make use of additional facts you know. These are
1/2 = 2^(-1); 16 = 2^4; so Log[1/2]/Log[16] = -1/4 when the base of each log is 2.
125 = 5^3; 25 = 5^2; so Log[125]/Log[25] = 3/2 when the base of each log is 5.
81 = 3^4; 27 = 3^3; so Log[81]/Log[27] = 4/3 when the base of each log is 3.
Now we can find x easily as the product of these fractions.
x = (-1/4)*(3/2)*(4/3) = (-1*3*4)/(4*2*3) = (-1/2)*(3*4)/(3*4)
x = -1/2
