SOLUTION: I tried these problems but I'm not sure if they're right. There were 3 parts, or 3 different equations to solve rather. part a) ln(x-4) + ln(2x) = ln(6). first i distributed the ln

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I tried these problems but I'm not sure if they're right. There were 3 parts, or 3 different equations to solve rather. part a) ln(x-4) + ln(2x) = ln(6). first i distributed the ln      Log On


   



Question 172872: I tried these problems but I'm not sure if they're right. There were 3 parts, or 3 different equations to solve rather. part a) ln(x-4) + ln(2x) = ln(6). first i distributed the ln to the (x-4) and came up with ln(x) - ln(4) + ln(2x) = ln(6). I dont know if this is possible, but then i canceled out all of the lns and was left with 3x=10, meaning x=10/3.
partb) e^2x-7 = 9. I know you have to take the ln of each side, but then i got stuck after i put the exponent as the coefficient leaving me with 2x-7lne = ln9
partc) log (9x+5) - log3 =2. not sure how to do this one.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Well, as they say in the old country..."you can't that there 'ere"
Let's see how it should go:
a) ln%28x-4%29%2Bln%282x%29+=+ln%286%29 Apply the "product rule" for logarithms...log%28M%29%2Blog%28N%29+=+log%28M%2AN%29
ln%28%28x-4%29%282x%29%29+=+ln%286%29 Simplify the left side.
ln%282x%5E2-8x%29+=+ln%286%29 ...and if log%28M%29+=+log%28N%29 then M+=+N, so...
%282x%5E2-8x%29+=+6 Subtract 6 from both sides.
2x%5E2-8x-6+=+0 Solve by the quadratic formula:
x+=+%28-%28-8%29%2B-sqrt%28%28-8%29%5E2-4%282%29%28-6%29%29%29%2F2%282%29
x+=+%288%2B-sqrt%2864%2B48%29%29%2F4
x+=+%288%2B4sqrt%287%29%29%2F4 or x+=+%288-2sqrt%287%29%29%2F4
x+=+2%2Bsqrt%287%29 or x+=+2-sqrt%287%29
-------------------------------------------
b) e%5E%282x-7%29+=+9 Take the natural log of both sides.
ln%28e%5E%282x-7%29%29+=+ln%289%29 Apply the "power rule" to the left side.
%282x-7%29ln%28e%29+=+ln%289%29 Substitute ln%28e%29+=+1
2x-7+=+ln%289%29 Evaluate the ln(9).
2x-7+=+2.197 Add 7 to both sides.
2x+=+9.197 Divide both sides by 2.
x+=+4.598
-------------------------------------------
c)log%28%289x%2B5%29%29-log%283%29+=+2 Apply the "quotient rule" to the left side.
log%28%289x%2B5%29%2F3%29+=+2 Rewrite in exponential form.
10%5E2+=+%289x%2B5%29%2F3 Multiply both sides by 3.
300+=+9x%2B5 Subtract 5 from both sides.
295+=+9x Divide both sides by 9.
x+=+32.77