algebra with logarithms question Watch

jojo55
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Hi everyone

I have the below question

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With the answer

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I understand most of the answer apart from how there is a 1 on the right hand side. (on the line listed as number 2).

Could someone please help explain what is going on?

Thanks
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ghostwalker
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(Original post by jojo55)
Could someone please help explain what is going on?

Thanks
ln t = 0 implies t = 1

e^0=1 so \ln 1 = 0

This is true of any log base.
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brianeverit
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(Original post by jojo55)
Hi everyone

I have the below question

Name:  tippy.JPG
Views: 88
Size:  21.3 KB

With the answer

Name:  tippy answer.JPG
Views: 100
Size:  41.9 KB

I understand most of the answer apart from how there is a 1 on the right hand side. (on the line listed as number 2).

Could someone please help explain what is going on?

Thanks
You can avoid the problem by noting that, using laws of logarithms gives
 \mathrm{ln} \frac{(y+1)(y-1)}{3}= \mathrm{ln} \frac{1}{3x+9} \to (y^2-1)= \frac{3}{3x+9}
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jojo55
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Report Thread starter 5 years ago
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(Original post by ghostwalker)
ln t = 0 implies t = 1

e^0=1 so \ln 1 = 0

This is true of any log base.

I didn't have that rule written down, but that now makes sense so I fully understand the question. Thanks for taking the time to help me
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