F(x)= 5^3x , the f^-1
(inverse) of (5 √5) is
i know the answer is 1/2 , but dont use the difficult way
like this
ln x = 3y ln(5) 3y = ln(x)/ln(5) 3y =1.4306765580733929 *log(x) y = 0.4768921860244643*l og(x) y = log(x)^ (0.4768921860244643) . ..................Ans put x =(5)^3/2 y = log{(5)^3/2}* (0.4768921860244643) y = log (5)^0.71533827903669 645 y = 0.71533827903669645 *log (5) =0.5000000000000000 1545 y = 1/2 ................Ans
(inverse) of (5 √5) is
i know the answer is 1/2 , but dont use the difficult way
like this
ln x = 3y ln(5) 3y = ln(x)/ln(5) 3y =1.4306765580733929 *log(x) y = 0.4768921860244643*l og(x) y = log(x)^ (0.4768921860244643) . ..................Ans put x =(5)^3/2 y = log{(5)^3/2}* (0.4768921860244643) y = log (5)^0.71533827903669 645 y = 0.71533827903669645 *log (5) =0.5000000000000000 1545 y = 1/2 ................Ans
-
i think its have true