Just got done taking a test in my Alg class. I had trouble with one problem though. 2^(x+1)=5^(3x) My answer came out to line -.66 when solved and he said to leave it in logarithmic form. My answer was something like x =((ln5/ln2)-1)/-2. I looked through my notes and everything and couldn't find a problem like it, can someone help me?
take logs of both sides.. (x+1)log2 = (3x)log 5 x+1 = 3x log5/log2 x+1 = (3x)(2.3219) x+1 = 6.9658x 5.9658x = 1 x = 1 / 5.9658 = 0.1676
That's much clearer -ln2/(ln2-3ln5). I can't believe I messed something up so easy lol. Oh well, should get some credit.
Too bad this was a math problem. Had it been an essay or thesis, all you would have needed to do was post it anywhere in the forum and the grammar and spelling police would have went over it with a fine tooth comb, essentially editing it for free.
I goy a ti-83plus because it was free through the school (16 in college, my high school pays for everything). What's a ti-89 do?
I hurried through mine w/o thinking all the way. I have a minor in Math so I tend to over-think it and solve too fast. I made a minor error in translating, here it is... 2^(x+1) = 5^(3x) log2^(x+1) = log 5^(3x) (x+1)log 2 = (3x)log 5 xlog2 + log2 = 3xlog5 xlog2 - 3xlog5 = -log2 x(log2 - 3log5) = -log2 x = -log2 / (log2 - 3log5) -this is the form your teacher was asking for... if you were to solve for x, then you would get .167 It has been a very long time since I have done these, but I am pretty confident this is correct.