SOLUTION: Hi, I'm having difficulty solving this equation. ln(2^(4x-1))=ln(8^(x+5))+logbase2(16^(1-2x)) (expressing the answer in terms of ln2) I simplified it until ln(2^(x-16))=4-8x W

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Hi, I'm having difficulty solving this equation. ln(2^(4x-1))=ln(8^(x+5))+logbase2(16^(1-2x)) (expressing the answer in terms of ln2) I simplified it until ln(2^(x-16))=4-8x W      Log On


   

Question 596631: Hi, I'm having difficulty solving this equation.
ln(2^(4x-1))=ln(8^(x+5))+logbase2(16^(1-2x))
(expressing the answer in terms of ln2)
I simplified it until
ln(2^(x-16))=4-8x
Where do I go from there?
Thanks
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
ln%282%5E%284x-1%29%29=ln%288%5E%28x%2B5%29%29%2Blog%282%2C+%2816%5E%281-2x%29%29%29%29
You have done a lot of the work already. Your equation:
ln%282%5E%28x-16%29%29=4-8x
is correct. The next step is to use a property of logarithms, log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29, to move the exponent of the argument out in front:
%28x-16%29ln%282%29=4-8x

Multiplying out the left side we get:
x%2Aln%282%29-16ln%282%29=4-8x

Next we gather all the terms with on in them on one side of the equation and the other terms on the other side. Adding 8x and 16ln(2) to each side we get:
x%2Aln%282%29+%2B+8x+=+16ln%282%29+%2B+4

Next we factor out x on the left side:
x%28ln%282%29+%2B+8%29+=+16ln%282%29+%2B+4
And last we divide both sides by (ln(2) + 8):
x+=+%2816ln%282%29+%2B+4%29%2F%28ln%282%29+%2B+8%29

And since the fraction on the right will not simplify, we are finished!