Asked by marcus
My book tells me to solve each exponential equation. I am given a lot of problems such as
6^x+1 = 4^2x-1
where I use In in the problem to help get my answer. However, in the same section, the book starts giving me problems such as
3(2)^x-2 +1=100
and .05(1.15)^x =5 etc.
Where I apparently have to use log in the equations instead of using In like I did in all the other problems. Why is this? Like, how will I know weather to use In or log to find my answer? Is there a reason why you use log instead?
6^x+1 = 4^2x-1
where I use In in the problem to help get my answer. However, in the same section, the book starts giving me problems such as
3(2)^x-2 +1=100
and .05(1.15)^x =5 etc.
Where I apparently have to use log in the equations instead of using In like I did in all the other problems. Why is this? Like, how will I know weather to use In or log to find my answer? Is there a reason why you use log instead?
Answers
Answered by
Kay
It makes no difference if you use ln or log unless the exercise tells you to. If you are allowed a calculator then it makes no difference.
In the example above:
3*2^(x-2)+1=100 3*2^(x-2)=99 2^(x-2)=33
If we use ln we get: ln2^(x-2)=ln33
(x-2)ln2=ln33 x= 2+ln33/ln2=7.044
If we used log with base 2 we would get: x-2=log(base2)33 so x= 2+log(base2)33 =7.044
No difference!
In the example above:
3*2^(x-2)+1=100 3*2^(x-2)=99 2^(x-2)=33
If we use ln we get: ln2^(x-2)=ln33
(x-2)ln2=ln33 x= 2+ln33/ln2=7.044
If we used log with base 2 we would get: x-2=log(base2)33 so x= 2+log(base2)33 =7.044
No difference!
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