Question 769968: I understand the concepts behind natural logarithms and solving for an exponent, but on this test review, there are problems for which I was not prepared by either my text or the homework. Here is an example:
x^2-6x+4=e^(x-7)+3
We are meant to use the calculator and solve for x as an approximate decimal. I am baffled as to what to do once I have gotten to
ln(x^2-6x+1)=(x-7)ln(e)
How do I solve for x when (x-7) is alone, (having been multiplied by ln(e), which is one)but the quadratic is still in ln form?
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! x^2-6x+4=e^(x-7)+3
We are meant to use the calculator and solve for x as an approximate decimal. I am baffled as to what to do once I have gotten to
ln(x^2-6x+1)=(x-7)ln(e)
How do I solve for x when (x-7) is alone, (having been multiplied by ln(e), which is one)but the quadratic is still in ln form?
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ln(e) = 1, so you can eliminate that.
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ln(x^2-6x+1)=(x-7)
That doesn't help a lot. The only method here is graphical or numerical.
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x =~ 0.1713815470435187
x =~ 5.885861739266182
x =~ 11.0354521340621
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