View Full Version : Math Wizards
Joshmil53
10-31-2012, 12:21 PM
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?
Phoenix
10-31-2012, 12:33 PM
42.....?
Joshmil53
10-31-2012, 12:35 PM
42.....?
I thought someone might say that lol
Phoenix
10-31-2012, 12:37 PM
I thought someone might say that lol
:laugh2::D
Joshmil53
10-31-2012, 12:44 PM
Why can't I edit threads after so long? A mod can delete/lock this please.
theogt
10-31-2012, 12:47 PM
Slightly embarrassed I don't remember how to solve this.
dexternjack
10-31-2012, 12:53 PM
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
Joshmil53
10-31-2012, 01:06 PM
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.
Diogenes
10-31-2012, 01:13 PM
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.
Yeagermeister
10-31-2012, 01:15 PM
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.
:laugh1:
Phoenix
10-31-2012, 01:21 PM
That's much clearer -ln2/(ln2-3ln5). I can't believe I messed something up so easy lol. Oh well, should get some credit.
If you say so. :write::ohno::huh::):o:
bigdnlaca
10-31-2012, 01:33 PM
Well I knew that it was higher than 0 but less than 1.
SaltwaterServr
10-31-2012, 01:54 PM
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?
Do yourself a HUGE favor and get a TI-89 Platinum if you're allowed to use it in class.
Joshmil53
10-31-2012, 02:00 PM
Do yourself a HUGE favor and get a TI-89 Platinum if you're allowed to use it in class.
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?
dexternjack
10-31-2012, 02:06 PM
That's much clearer -ln2/(ln2-3ln5). I can't believe I messed something up so easy lol. Oh well, should get some credit.
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.
Joshmil53
10-31-2012, 02:13 PM
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.
It is.
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