1) log3 27-log1/7 7 2) 2^1+log2 5 3) lg 4+2 lg 5 4) найдите значение х если a) 3^x=7 b) log4 x=log0,5 корень 2

1)

log3 (27) - log (1/7) (7) = log3 (3³) - log7 (7) / log7 (1/7) = 3log3 (3) - 1/log7 (7^ (-1)) = 3 - 1 / (-1) = 3 + 1 = 4.

2)

2^ (1 + log2 (5)) = 2^ (log2 (2) + log2 (5)) = 2^log2 (2 * 5) = 2^log2 (10) = 10.

3)

lg4 + 2lg (5) = lg4 + lg (5²) = lg4 + lg (25) = lg (4 * 25) = lg100 = lg (10²) = 2.

4)

a)

3^x = 7;

log3 (3^x) = log3 (7);

xlog3 (3) = log3 (7);

x = log3 (7).

b)

log4 (x) = log0,5 (√2);

log2 (x) / log2 (4) = log2 (2^ (1/2)) / log2 (1/2);

log2 (x) / log2 (2²) = (1/2) / log2 (2^ (-1/2));

log2 (x) / 2 = (1/2) / (-1/2);

log2 (x) / 2 = - 1;

log2 (x) = - 2;

x = 2^ (-2) = 1/4.