Question

log b3 a2 × log c3 b2 log a3 c2 = 8 / 27​

Answer 1

Step-by-step explanation:

$log_ {{b}^{3}} {a}^{2} \times log_ {{c}^{3}} {b}^{2} \times log_ {{a}^{3}} {c}^{2} = \frac{8}{27} \\\\ LHS = log_ {{b}^{3}} {a}^{2} \times log_ {{c}^{3}} {b}^{2} \times log_ {{a}^{3}} {c}^{2} \\\\ by \: change \: of \: base \: law : \\ \\= \frac{log {a}^{2} }{log {b}^{3} } \times \frac{log {b}^{2} }{log {c}^{3} } \times \frac{log {c}^{2} }{log {a}^{3} } \\ \\= \frac{2log {a} }{3log {b} } \times \frac{2log {b} }{3log {c}} \times \frac{2log {c} }{3log {a} } \: \\ \\ = \frac{(2 \times 2 \times 2) \times loga \: logb \: log \: c}{(3 \times 3 \times 3)\times loga \: logb \: log \: c} \\ \\ = \frac{8}{27} \\ \\= RHS$

Thus proved.

Answer 2

Step-by-step explanation:

as log(base a)b = logb/loga

so, loga²/logb³×logb²/logc³×logc²/loga³=8/27

LHS,,

2/3×2/3×2/3 = 8/27 = RHS

hence proved...