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How do you solve this equation for the variable? 3^2m= 81

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rosey-girl | Student, Grade 12 | Valedictorian

Posted May 3, 2013 at 9:28 PM via web

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How do you solve this equation for the variable?

3^2m= 81

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rosey-girl | Student, Grade 12 | Valedictorian

Posted May 3, 2013 at 9:32 PM (Answer #1)

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using natural log

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oldnick | Valedictorian

Posted May 3, 2013 at 10:09 PM (Answer #2)

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`3^(2m)=3^4`

`2m=4`

`m=2`

using natural log:

`2m ln3= 4ln 3`

`2m=4`

`m=2`

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violy | High School Teacher | (Level 3) Assistant Educator

Posted May 4, 2013 at 12:00 AM (Answer #4)

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We can first rewrite 81 as an exponential with base 3. 

We know that 3*3*3*3 = 81. So, 81 = 3^4.

We will have:

3^2m = 3^4

Using natrual logarithm on both sides. 

ln3^2m = ln3^4

Use the property: lna^b = blna.

2mln3 = 4ln3

Divide both sides by ln3. 

2m = 4

Divide both sides by 2. 

m = 2

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pramodpandey | College Teacher | Valedictorian

Posted May 4, 2013 at 6:25 AM (Answer #5)

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3^2m= 81

`3^(2m)=81`

`(3^m)^2=9^2`

`(3^m)^2-9^2=0`

Factor above expression by `A^2-B^2=(A-B)(A+B)`

`(3^m-9)(3^m+9)=0`

`3^m-9=0`

`because 3^m+9!=0AA m`

`3^m=3^2`

`m=2`

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rosey-girl | Student, Grade 12 | Valedictorian

Posted May 3, 2013 at 10:28 PM (Answer #3)

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this is not how you do it, sorry but my teacher told me to use natural log or log base.

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