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
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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
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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.
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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.