Question

log25-log9/log125-log27

Answer 1

$\bold{\frac{log25 - log9}{log125-log27}}\\\\=\bold{\frac{log5^2-log3^2}{log5^3-log3^3}}$
we know,
Logaⁿ = nloga , use this concept here ,
$=\bold{\frac{2log5-2log3}{3log5-3log3}}\\\\=\bold{\frac{2(log5-log3)}{3(log5-log3)}}\\\\=\bold{\frac{2}{3}}$

Hence answer is 2/3

Answer 2

Hello Dear.

Here is your answer---



Given---

  (log 25 - log 9) ÷ (log 125 - log 27)
=  (log 5² - log 3²) ÷ (log 5³ - log 3³)

We know, log aⁿ = n log a

Thus, $\frac{2 log 5 - 2log 3}{3 log5 - 3 log 3}$
       =  $\frac{2(log5 - log 3)}{3 (log5 - log 3)}$
        = 2/3

Thus, the answer of the Above required question will be 2/3.


Hope it helps.


Have a Marvelous Day.