BethArkless
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Hello! New to this so not too sure how to work it was just wondering if any of you clever students out there could help me with this question:
Solve for x, round your answers to 2 dec places,
ln(x) + ln(4x) - ln(2x) = 1
Hope someone can help! Thank you!
Beth.
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Cyclohexane
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Use your laws of logs to get the left hand side as a single logarithm:

loga + logb = log(ab)
loga - logb = log(a/b)


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BethArkless
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Okay, what's the inverse of ln?
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Ollie113
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(Original post by BethArkless)
Hello! New to this so not too sure how to work it was just wondering if any of you clever students out there could help me with this question:
Solve for x, round your answers to 2 dec places,
ln(x) + ln(4x) - ln(2x) = 1
Hope someone can help! Thank you!
Beth.
(Original post by Cyclohexane)
Use your laws of logs to get the left hand side as a single logarithm:

loga + logb = log(ab)
loga - logb = log(a/b)


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Simples.

Use the rules quoted to get ln((4x^2)/2x)=1
Simplify inside the brackets to get ln(2x)=1
Now for the part that you either undersrstand or don't, if you don't understand just accept this as maths magic and learn where it applies: You are using ln, the natural log of the number e
Therefore ln(e)=1
And e*ln(x)=x
Therefore if we multiply both sides by e we get:
2x=e
Therefore our answer is x=(1/2)e
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Robbie242
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(Original post by Ollie113)
Simples.

Use the rules quoted to get ln((4x^2)/2x)=1
Simplify inside the brackets to get ln(2x)=1
Now for the part that you either undersrstand or don't, if you don't understand just accept this as maths magic and learn where it applies: You are using ln, the natural log of the number e
Therefore ln(e)=1
And e*ln(x)=x
Therefore if we multiply both sides by e we get:
2x=e
Therefore our answer is x=(1/2)e
I swear that's like C3/C4 level? all the OP has to do is start by using the multiplication law and then the division law to express it is a single logarithm. Unless this is C3/C4 level of course
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Ollie113
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(Original post by Robbie242)
I swear that's like C3/C4 level? all the OP has to do is start by using the multiplication law and then the division law to express it is a single logarithm
Nah man easier than that this is C2 stuff, and yeah well easy. But I'm guessing the OP was just confused as to what the inverse of ln was, perhaps her maths teacher forgot to explain the ln and e are inverses of eachother.
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BethArkless
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Would you not just bring ln to the other side to make it 2x=e^1 ? And then just divide tat answer by 2 to find x?
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Robbie242
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(Original post by BethArkless)
Would you not just bring ln to the other side to make it 2x=e^1 ? And then just divide tat answer by 2 to find x?
Just an assumption here, but I swear anything to the power of 1 is just the integer/algebraic value, so its exactly the same as what Ollie113 said, divide it by 2 and you get \frac{1}{2}e
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BethArkless
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Okay, thanks for your help guys
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williamnguyen
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ln(x) + ln(4x) - ln(2x) = 1

ln (4x^2) - ln (2x) = 1
ln (2x) = 1
2x = e
x = e/2
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