How do you solve this equation for the variable?
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using natural log
using natural log:
`2m ln3= 4ln 3`
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
Factor above expression by `A^2-B^2=(A-B)(A+B)`
`because 3^m+9!=0AA m`
this is not how you do it, sorry but my teacher told me to use natural log or log base.
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