Can Logarithms Simplify Exponential Equations?

[PrudensOptimus]

IS there a way to find x if given:

3^x - 3^(x-1) = 1000


I tried to take log [3^x - 3^(x-1)]/log[3] = log1000/log3

but then, x-x-1 = ...

 

[Warr]

well, I've never done logarithms really, but I figured out how to do this


to break it down it goes like this

(3*3*3*3...*3*3) - (3*3*3*3..*3) = 1000

so therefore 1 less 3 is multiplied in the second part...heres a simple example so you can see the pattern

(5*5*5*5) - (5*5*5) = (5*5*5)*(5-1)

therefore we can say in general that this means

N^m - N^(m-1) = (N^(m-1))*(N-1)

now instead of the left side of the equation above, we use the right side, and solve for x

(3^(x-1))*(3-1) = 1000
3^(x-1) = 1000/(3-1)
3^(x-1) = 500
log (3^(x-1)) = log (500)
(x-1)(log 3) = log (500)
x-1 = log (500)/log (3)
x = [log (500)/log (3)] + 1

and voila, you have x...

another way of looking at this is that since you are subtracting something with one less *3, then 3^x : 3^(x-1) is 3:1. 3-1=2, so 2:1000. This means that 3:1500 and 1:500. Therefore 3^x = 1500 and x^(x-1) = 500

I hope this helps

 

[Integral]

You can use the properties of exponents to factor your inital expression.

3^x - 3^{(x-1)}=1000
3^x -3^x 3^{-1}=1000
3^x(1 - 3^{-1})=1000
3^x= 1500
ln(3^x)= ln 1500
x ln 3= ln 1500
x = \frac {ln 1500} {ln 3}

I'll leave it to the reader to show that the 2 results are the same.

 

[oen_maclaude]

another solution:
instead of factoring 3^x, factor out 3^x-1
(3^x)-{3(x-1)}=1000
{3^(x-1)}[3-1]=1000
{3^(x-1)}2=1000
{3^(x-1)}=500
log{3^(x-1)}=log(500)
(x-1)log3=log500
x=[(log500)/(log3)]+1

 

[evagriselda]

well, I've never done logarithms really, but I figured out how to do this


to break it down it goes like this

(3*3*3*3...*3*3) - (3*3*3*3..*3) = 1000

so therefore 1 less 3 is multiplied in the second part...heres a simple example so you can see the pattern

(5*5*5*5) - (5*5*5) = (5*5*5)*(5-1)

therefore we can say in general that this means

N^m - N^(m-1) = (N^(m-1))*(N-1)

now instead of the left side of the equation above, we use the right side, and solve for x

(3^(x-1))*(3-1) = 1000
3^(x-1) = 1000/(3-1)
3^(x-1) = 500
log (3^(x-1)) = log (500)
(x-1)(log 3) = log (500)
x-1 = log (500)/log (3)
x = [log (500)/log (3)] + 1

and voila, you have x...

another way of looking at this is that since you are subtracting something with one less *3, then 3^x : 3^(x-1) is 3:1. 3-1=2, so 2:1000. This means that 3:1500 and 1:500. Therefore 3^x = 1500 and x^(x-1) = 500

I hope this helps

How can I understand the logarithms requirements in flowcharts? Easily please, I am mad about a situation with a proffesor.

 

[PrudensOptimus]

How can I understand the logarithms requirements in flowcharts? Easily please, I am mad about a situation with a proffesor.

u mean something like

log (a*b) = log a + log b?

they are called Log Rules or Log Properties.