Equazione logaritmica n. 528

@Beppe

Ciao Beppe, 

Utilizziamo la proprietà dei logaritmi del cambio di base. 

Insieme di definizione in R: x>0

Buona giornata.

Stefano 

Ciao @beppe

Sempre con i soliti problemi...

Cambio base: e (logaritmi neperiani)

riscrivo:

LN(x^2 + 1)/LN(2) = 1 + 2·LN(x)/(3·LN(2)) + LN(x)/(3·LN(2))

LN(x^2 + 1)/LN(2) = LN(2)/LN(2) + 2·LN(x)/(3·LN(2)) + LN(x)/(3·LN(2))

C.E. x > 0

moltiplico per LN(2)

LN(x^2 + 1) = LN(2) + (2·LN(x)/3 + LN(x)/3)

LN(x^2 + 1) = LN(2) + LN(x)

LN(x^2 + 1) = LN(2·x)

Quindi:

x^2 + 1 = 2·x-----> x^2 - 2·x + 1 = 0----> (x - 1)^2 = 0

x=1 soluzione

regola :

log(2) 10 = log(8) 10 * log(2) 8

3,321928 = 1,107309 * 3,000000

3,321928 = 3,321928...it works !!!

Se il problema é il log8 x ecco qua

log_2(x) = log x/log 2 = 3log x/(3log 2)=

=log x^3 /log2^3 = log_8 (x^3 ). 

log_2 (x^2 + 1) = 1 + 2/3 log_2 (x) + log_8 (x)

Posto x > 0

e 1 = log_2 (2)

log_8 (x) = log(x)/log(8) = log(x)/log(2^3) = log(x)/(3 log(2)) = 1/3 * log(x)/log(2) = 1/3 log_2(x)

risulta

log_2(x^2 + 1) = log_2(2) + 2/3 log_2(x) + 1/3 log_2(x)

log_2(x^2+1) = log_2(2) + log_2(x)

log_2(x^2 + 1) = log_2 (2x)

x^2 + 1 = 2x

x^2 - 2x + 1 = 0

(x - 1)^2 = 0

x - 1 = 0

x = 1

* log(base, argomento) = log(b, a) = ln(a)/ln(b)
* log(b, a^n) = n*ln(a)/ln(b) = n*log(b, a)
* log(b^n, a) = ln(a)/(n*ln(b)) = (1/n)*log(b, a)
------------------------------
528) log(2, x^2 + 1) = 1 + (2/3)*log(2, x) + log(8, x) ≡
≡ log(2, x^2 + 1) = 1 + (1/3)*log(2, x^2) + (1/3)*log(2, x) ≡
≡ log(2, x^2 + 1) = 1 + (1/3)*log(2, x^3) ≡
≡ log(2, x^2 + 1) = 1 + log(2, x) ≡
≡ log(2, x^2 + 1) - log(2, x) = 1 ≡
≡ log(2, (x^2 + 1)/x) = 1 ≡
≡ 2^log(2, (x^2 + 1)/x) = 2^1 ≡
≡ (x^2 + 1)/x = 2 ≡
≡ x^2 - 2*x + 1 = 0 ≡
≡ (x - 1)^2 = 0 ≡
≡ x = 1