Question

Log 729 по основанию 2log 9 по основанию 2​

Answer 1

$log_{2}(729) = log_{2}( {3}^{6} ) = 6 log_{2}(3)$

$log_{2}(9) = log_{2}( {3}^{2} ) = 2 log_{2}(3)$

Answer 2

To simplify the expression "log 729 base 2 / log 9 base 2," we can use the logarithm identity:

log_a(b) = log_c(b) / log_c(a)

where "a" is the base of the logarithm and "c" is any positive number.

So, we have:

log 729 base 2 / log 9 base 2 = log_c(729) / log_c(9)

We can choose any positive number "c" for the logarithm identity, as long as it is the same for both the numerator and the denominator. Let's use "c = 10":

log 729 base 2 / log 9 base 2 = log_10(729) / log_10(9)

Now, calculate the logarithms:

log_10(729) = 2.8627 log_10(9) = 0.9542

So, the final simplified expression is:

log 729 base 2 / log 9 base 2 ≈ 2.8627 / 0.9542 ≈ 2.9998

Therefore, log 729 base 2 / log 9 base 2 is approximately equal to 2.9998.