Question

(vii) log2 log,1/3 log8(x - 1)>0​

Answer 1

your question is -> $log_2{log_{1/3}{log_{8}{(x-1)}}}$ > 0

we know, loga N = x ⇒N = a^x

so, log2{log(1/3)log8 (x - 1)} > 0

⇒log(1/3) log8 (x - 1) > 2^0

⇒log(1/3) log8 (x - 1) > 1

as base of logarithm lies between 0 to 1

so, sign of inequality will change.

⇒log8 (x - 1) < (1/3)¹

⇒log(2³) (x - 1) < 1/3

we know, logaⁿ N = (1/n) loga N

⇒(1/3) log2 (x - 1) < (1/3)

⇒log2 (x - 1) < 1

⇒(x - 1) < 2¹

⇒x < 2 + 1 = 3

hence, x < 3......(1)

for log to be defined, (x - 1) > 0

so, x > 1 .......(2)

for log to be defined , log8 (x - 1) > 0

x - 1 > 8^0 ⇒x > 2 .......(3)

also for log to be defined, log(1/3)log8 (x - 1) > 0

or, log8 (x - 1) < 1

or, x - 1 < 8

or, x < 9 ........(4)

from equations (1), (2), (3) and (4), we get 2 < x < 3

hence, value of x ∈ (2, 3)