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

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) (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) Apply the "product rule" for logarithms...
Simplify the left side.
...and if then , so...
Subtract 6 from both sides.
Solve by the quadratic formula:

or
or
-------------------------------------------
b) Take the natural log of both sides.
Apply the "power rule" to the left side.
Substitute
Evaluate the ln(9).
Add 7 to both sides.
Divide both sides by 2.

-------------------------------------------
c) Apply the "quotient rule" to the left side.
Rewrite in exponential form.
Multiply both sides by 3.
Subtract 5 from both sides.
Divide both sides by 9.

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