Exponents as VariablesDate: 11/20/2001 at 16:27:49 From: Bart Subject: Exponents as a variable I can solve these problems but I don't know how to prove them (or logically show my work): 9^1+x = 27 2^-x-4 = 1/32 243 = (1/3)^x+4 64 = 0.5)^3-x I can solve for x but I can't show how to isolate x in order to show how I solved it. Can you please help me? Thanks, Bart Date: 11/20/2001 at 17:08:27 From: Doctor Peterson Subject: Re: Exponents as a variable Hi, Bart. I'd like to hear how you solved these; probably that can be written down in a reasonable way. There are two ways I can see to approach these problems. One is to take the log of both sides: 9^(1+x) = 27 (1+x) log(9) = log(27) 1+x = log(27)/log(9) x = log(27)/log(9) - 1 Now, how can you simplify this? One way is to carefully choose what base to use for the log; since both 27 and 9 are powers of 3, with 9 = 3^2 and 27 = 3^3, if you use the base 3 log you get x = 3/2 - 1 = 1/2 Or, you could just express 9 and 27 as powers and simplify: x = log(3^3)/log(3^2) - 1 3 log(3) = -------- - 1 = 3/2 - 1 = 1/2 2 log(3) The other way to approach these problems is to write 9 and 27 as powers from the start, and not use logarithms explicitly at all: 9^(1+x) = 27 (3^2)^(1+x) = 3^3 3^(2+2x) = 3^3 2+2x = 3 (equating the exponents) 2x = 1 x = 1/2 The step where I equated the exponents is really the same as taking the base 3 logarithm of both sides. One of those may be more or less what you are doing mentally. Choose whichever way feels most comfortable to you, and try the other problems. Let me know if you need any more help. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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