Stuck on a Log :(

**BrendanS** · January 21st 2008, 09:28 AM

Here is what i have thus far..

2^(3x-4) -3^(2x+5) = 7
3xlog2 -4log2 +2xlog-3 +5log-3 = 7
3xlog2 +2xlog-3 = 7 + 4log2 -5log-3
3xlog2+ 2xlog-3 = 7 +log256 - log 243
3xlog2+ 2xlog-3 = 7 + log(256/243)
3xlog2+ 2xlog-3 = 8.05

Not sure where to take it from here. Any help is appreciated <3.

**colby2152** · January 21st 2008, 09:38 AM

You calculated the log values on the right hand side, so why can't you do it on the left hand side? Also, you never took the log of the seven in the first line. BTW, you cannot take the log of a negative number...

$2^{(3x-4)} -3^{(2x+5)} = 7$

$(3x-4)log(2) -(2x+5)log(3) = log(7)$

Calculate the value of those logs, and then solve for x.

**BrendanS** · January 21st 2008, 09:43 AM

Ah, thanks. Yea.. im not actually sure how to isolate the x's on the left side and I cant find my note on it :|. Ive got it down to..

3xlog2 -2xlog3 = 7.37

**colby2152** · January 21st 2008, 09:47 AM

Originally Posted by BrendanS

Ah, thanks. Yea.. im not actually sure how to isolate the x's on the left side and I cant find my note on it :|. Ive got it down to..

3xlog2 -2xlog3 = 7.37

$log(2)=.301$ , $log(3)=.477$

$.903x-.954x=7.37$

$x=-143.83$

This is correct given that the right side (7.37) was correctly done.

**BrendanS** · January 21st 2008, 09:56 AM

Thanks your a life saver <3, last day before exams start and im the only one that showed up for class with a single question >_> .

**Krizalid** · January 21st 2008, 10:39 AM

Originally Posted by colby2152

$2^{(3x-4)} -3^{(2x+5)} = 7$

$(3x-4)log(2) -(2x+5)log(3) = log(7)$

Careful with this step.

Recall $\log(a-b)$ has not property.

**colby2152** · January 21st 2008, 11:07 AM

Originally Posted by Krizalid

Careful with this step.

Recall $\log(a-b)$ has not property.

What do you mean? Yes, you cannot take the log of the negative number, but you can switch it to the right side and do the same calculation.

**Krizalid** · January 21st 2008, 11:29 AM

Originally Posted by colby2152

What do you mean? Yes, you cannot take the log of the negative number

I never spoke about logs of negative numbers.

---

What you did here

Originally Posted by colby2152

$2^{(3x-4)} -3^{(2x+5)} = 7$

$(3x-4)log(2) -(2x+5)log(3) = log(7)$

It is incorrect, 'cause $\log (a - b) \ne \log a - \log b.$

---

I suggest to make a graph of the equation.

**colby2152** · January 21st 2008, 11:46 AM

True true.. brain fart there, thought I could do that because I am so used to working with $Ae^x=B$ forms.

**earboth** · January 21st 2008, 12:10 PM

Originally Posted by BrendanS

Here is what i have thus far..

2^(3x-4) -3^(2x+5) = 7
3xlog2 -4log2 +2xlog-3 +5log-3 = 7
3xlog2 +2xlog-3 = 7 + 4log2 -5log-3
3xlog2+ 2xlog-3 = 7 +log256 - log 243
3xlog2+ 2xlog-3 = 7 + log(256/243)
3xlog2+ 2xlog-3 = 8.05

Not sure where to take it from here. Any help is appreciated <3.

Hi,

$2^{3x-4} -3^{2x+5} = 7~\iff~\frac1{16} \cdot 8^x - 243 \cdot 9^x = 7$

By comparison of the coefficients and the bases it follows that the LHS is negative. Thus this equation doesn't have a real solution.

Thread: Stuck on a Log :(

LinkBack

Thread Tools

Display

Stuck on a Log :(

Similar Math Help Forum Discussions

Stuck

stuck with sin^4(x)

somebody help me plz im so stuck

stuck

Stuck, Stuck, Stuck - Need Help Urgently

Search Tags