# Math Help - Log or ln

1. ## Log or ln

My problem is 2^3x=3^x+4

i have been trying this problem and have done or gotten this far.. i dont know if its right or wrong but....

3x ln2= x+4 ln3

3 ln2 + x ln2 = x ln 3 + 4 ln 3

x ln 2 - x ln 3 = 4 ln 3 - 3 ln 2

x ln 2/3 = ? ln 3/2

this is what i got and im stumped am i on the right track and or what is the next step..

any help is appreciated

2. ## Re: Log or ln

$\displaystyle x (\ln 2 - x \ln 3) = 4 \ln 3 - 3 \ln 2$

$\displaystyle x = \frac{4 \ln 3 - 3 \ln 2}{\ln 2 - x \ln 3}$

And that's it!

3. ## Re: Log or ln

2^(3x) = 3^(x+4)
taking log on both sides we get
3x log 2 = ( x+4) log 3
3xlog2 - xlog 3 = 4log 3
x[ 3log 2 - log 3] = 4 log 3
x log ( 8/3) = 4log 3
Now you can workout the value of x

4. ## Re: Log or ln

Originally Posted by gramgilb
My problem is 2^3x=3^x+4

i have been trying this problem and have done or gotten this far.. i dont know if its right or wrong but....

3x ln2= x+4 ln3

3 ln2 + x ln2 = x ln 3 + 4 ln 3

x ln 2 - x ln 3 = 4 ln 3 - 3 ln 2

x ln 2/3 = ? ln 3/2

this is what i got and im stumped am i on the right track and or what is the next step..

any help is appreciated
What you have posted is meaningless; If you are going to post in ASCII learn to use parentheses to make your expressions unambiguous.

The helpers have assumed you mean:

$2^{3x}=3^{x+4}$

which is not unreasonable given what follows, but that is not what you have posted means under the usual conventions which may be:

$2^{3x}=3^x +4$

or even

$2^3 x=3^x+4$

etc..

.

5. ## Re: Log or ln

Originally Posted by zzephod
What you have posted is meaningless; If you are going to post in ASCII learn to use parentheses to make your expressions unambiguous.

The helpers have assumed you mean:

$2^{3x}=3^{x+4}$

which is not unreasonable given what follows, but that is not what you have posted means under the usual conventions which may be:

$2^{3x}=3^x +4$

or even

$2^3 x=3^x+4$

etc..

.
K sorry didnt know that but will learn to do that for any other inquiries

thank you