4.42 ?: zadanie 4.42. log(2^x−1)−log2=log(2^x+3)−log(2^x−1) 2* log(2^x−1)=log(2^x+3)+log2 x∊D ⇔ 2^x−1>0 ⇒ x>0 ⋀ 2^x+3>0 ⇒ x∊R D=(0,∞) y=2^x−rosn. log(2^x−1)²=log(2^x+3)*2 f. różn. (2^x−1)²=(2^x+3)*2 i ozn. 2^x=p (p−1)²=(p+3)*2 p²−4p−5=0 √Δ=6 [p=−1 v p=5] i p=2^x −1=2^x 5=2^x ⇒log₂5=x prosze x∊Φ

?: to x∊Φ powinno byc pod tym zapisem −1=2^x