Math Wizards

[Joshmil53]

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]

42.....?

[Joshmil53]

Originally Posted by Phoenix

42.....?

I thought someone might say that lol

[Phoenix]

Originally Posted by Joshmil53

I thought someone might say that lol

[Joshmil53]

Why can't I edit threads after so long? A mod can delete/lock this please.

[theogt]

Slightly embarrassed I don't remember how to solve this.

[dexternjack]

Originally Posted by Joshmil53

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]

Originally Posted by dexternjack

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]

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]

Originally Posted by Diogenes

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.

[Phoenix]

Originally Posted by Joshmil53

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.

[bigdnlaca]

Well I knew that it was higher than 0 but less than 1.

[SaltwaterServr]

Originally Posted by Joshmil53

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]

Originally Posted by SaltwaterServr

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]

Originally Posted by Joshmil53

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.