Question

If logs 64 +logs 1/125 + 2 log2 8= x, what is the value of x?​

Answer 1

.

64+ 1/125 + 2+ 28 = x

64 + 1 +2 +28 = 125*x

67+28 = 125x

95 = 125x

125x = 95

x = 95/125

x= 19/25

Answer 2

Given:

log64 +log1/125 + 2 log2 8= x

To Find:

value of x

Solution:

According to logarithmic rules

loga + logb + logc = log(a×b×c)

alogb= log$b^{a}$

log(a/b) = loga - log b

So,

log64 +log1/125 + 2 log28 =  log64 +log1/125 + log(28)²

                                             =  log64 + log1 - log 125 + log(28)²

                                             =  log$2^{6}$ +log1 - log$5^{3}$ + log(28)²

                                             = 6×log2 + log1 - (3×log5) + 2×log28

                                             = (6×0.301)+0-(3×0.69)+(2×1.44)

                                             = 1.806-2.07+2.88

                                             =2.616

Hence the value of x is 2.616