SOLUTION: I just wanted to be positive I've got this right. I have a problem: (6)ln(x)=-2.78 My process is: 6e^-2.78=x - 6(0.06203850738) - 0.37. Am I right? If not, what is the correct proc
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-> SOLUTION: I just wanted to be positive I've got this right. I have a problem: (6)ln(x)=-2.78 My process is: 6e^-2.78=x - 6(0.06203850738) - 0.37. Am I right? If not, what is the correct proc
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Question 1103355: I just wanted to be positive I've got this right. I have a problem: (6)ln(x)=-2.78 My process is: 6e^-2.78=x - 6(0.06203850738) - 0.37. Am I right? If not, what is the correct process? Thank you! Answer by greenestamps(13215) (Show Source):
With the way you show your "process", I have no idea what you really did to try to solve this equation. Furthermore, I don't see where you have shown your final answer.
I recognize the number 0.06203850738 that you show in your work; it is . But I can't make any sense out of what you did with that number.
That number will of course show up in your calculations if you change the log equation to an exponential equation:
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If you are going to solve the equation that way, then you need to simplify the expression on the left hand side:
Then your equation is
There is a much simpler path to this answer.
The "6" is just a constant multiplier of ln(x). The easiest thing to do is first divide both sides of the equation by 6. Then the path to the answer is simple:
