Log9 (x + 1) + log9 (9x + 9) = 2pd : los 9 son la base?
Log9 (x + 1) + log9 (9x + 9) = 2 pd : los 9 son la base.
Log9 (x + 1) + log9 (9x + 9) = 2 pd : los 9 son la base.
En resumen
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Log en baseN de M = x de donde : M = N ^ x log de x en base 2 = - 1 / 2 x = 2 ^ - 1 / 2 x = raiz de 2 sobre 2 log de x en base 0. 3 = - 2 x = (3 / 10) ^ - 2 x = 100 / 9 log de x en base p = - 3 x = p ^ - 3 x = 1 / p ^ 3…
Log(ab) = log a + log b = 3 + 4 = 7 log a / b = log a - log b = 3 - 4 = - 1 log a ^ 2 = 2 log a = 2 (3) = 6 log ab este es igual a primero Suerte pd : El punto uno es asi log(a * b) o que quieres decir con log(a' * b).
4, 39 4, 39 4, 39 4, 39 4, 39.
Veamos : del dato, x >0 Log (x - 1) + Log x = Log 10 Log (x - 1)(x) = Log 10 x(x - 1) = 10 x² - x - 10 = 0 x = 1 + - √1² - 4 * 1 * ( - 10) / 2 * 1 x = 1 + - √41 / 2 x₁ = 1 + √41 / 2≈ 3. 7.
A) loga(36) = loga(6²) = 2loga(6) = 2loga(3x2) = 2[loga(3) + loga(2)] = 2 (y + z ) = 2y + 2z b) loga(72) = loga(9x8) = loga(9) + loga(8) = loga(3²) + loga(2³) = 2loga(3) + 3loga(2) = 2y + 3z.