Solve 3^x = 2^(x+1) Should I do this by equating indices?
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,582 Sep 27, 2006 #2 Since the bases are not the same, you can make no assumptions regarding the equality of the powers. Have you not studied logarithms yet? Thank you. Eliz.
Since the bases are not the same, you can make no assumptions regarding the equality of the powers. Have you not studied logarithms yet? Thank you. Eliz.
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,582 Sep 28, 2006 #4 Denis said: hint: 2^(x+1) = 2^x 2^1 Click to expand... Yes, but wouldn't logarithms still be necessary? It's not like the solution is a rational number.... Eliz.
Denis said: hint: 2^(x+1) = 2^x 2^1 Click to expand... Yes, but wouldn't logarithms still be necessary? It's not like the solution is a rational number.... Eliz.
D Denis Senior Member Joined Feb 17, 2004 Messages 1,707 Sep 28, 2006 #5 stapel said: Denis said: hint: 2^(x+1) = 2^x 2^1 Click to expand... Yes, but wouldn't logarithms still be necessary? It's not like the solution is a rational number.... Click to expand... Of course. 3^x = 2 * 2^x (3/2)^x = 2 ... was trying to let him know how to put it in this form, in order to use logs
stapel said: Denis said: hint: 2^(x+1) = 2^x 2^1 Click to expand... Yes, but wouldn't logarithms still be necessary? It's not like the solution is a rational number.... Click to expand... Of course. 3^x = 2 * 2^x (3/2)^x = 2 ... was trying to let him know how to put it in this form, in order to use logs
