Intersecting Functions: Solving Logarithmic Equations

  • Thread starter Thread starter TheRedDevil18
  • Start date Start date
  • Tags Tags
    Logarithmic
Click For Summary
To find the intersection points of the functions f(x) = 3^x and g(x) = 2^(x^2), the correct approach involves equating their logarithmic forms: log(3^x) = log(2^(x^2)). This leads to the equation x log(3) = x^2 log(2). The solution can be simplified to x = log(3)/log(2), with x = 0 also being a valid solution. The base of the logarithm is not fixed; any base will yield the same ratio for the logs. Understanding these transformations is crucial for solving logarithmic equations effectively.
TheRedDevil18
Messages
406
Reaction score
2

Homework Statement



Determine at which points the graphs of the given pair of functions intersect:

f(x) = 3x and g(x) = 2x2

Homework Equations





The Attempt at a Solution



I know I have to equate and solve for x so I converted them to logarithms

log3x = log2x2

Don't know if that's right, but I am stuck here, do I use the change of base formula ?
 
Physics news on Phys.org
TheRedDevil18 said:

Homework Statement



Determine at which points the graphs of the given pair of functions intersect:

f(x) = 3x and g(x) = 2x2

Homework Equations





The Attempt at a Solution



I know I have to equate and solve for x so I converted them to logarithms

log3x = log2x2

Don't know if that's right, but I am stuck here, do I use the change of base formula ?

With that subscript in there not clear what you mean by "converted them to logarithms". The correct thing to do is to try to solve the equation log(f(x))=log(g(x)). ##\log 3^x=\log 2^{x^2}##. The 'log' can be any base you like. Just use the rules of logarithms to solve that equation.
 
You cannot just replace exponentials by logarithms, it won't work. There is a way to solve it, but then your steps have to be valid transformations.
do I use the change of base formula ?
That is a good idea, you can do it with the exponentials as well.
 
log3x = log2x2

log3x = 2log2x

Using the change of base formula

log2x/log23 = 2log2x...stuck here
 
2x2 means 2(x2), not (2x)2, your first step does not work.

What is log(3x) simplified?
 
xlog3 = x^2log2

log3 = xlog2

x = log3/log2

Is that correct ?, also why is the base 10 ?, I thought it was 3 and 2 respectively
 
TheRedDevil18 said:
xlog3 = x^2log2

log3 = xlog2

x = log3/log2

Is that correct ?, also why is the base 10 ?, I thought it was 3 and 2 respectively

That's part of it. The base doesn't have to be 10. If you take ratio log(3)/log(2) in any base you'll get the same number. Can you say why? More importantly, there is another solution. What is it?
 
Dick said:
That's part of it. The base doesn't have to be 10. If you take ratio log(3)/log(2) in any base you'll get the same number. Can you say why? More importantly, there is another solution. What is it?

I think the other solution should be x = 0 as well ?, I'm not too sure about why you get the same number, a bit confused, can you explain that please ?
 
Starting from 3^x= 2^{x^2}, you can take the logarithm to any base, "10", "e", whatever, and get log(3^x)= x log(3)= log(2^{x^2})= x^2 log(2). If x is not 0, you can divide both sides by x log(2).
 

Similar threads

Replies
4
Views
2K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 26 ·
Replies
26
Views
1K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 11 ·
Replies
11
Views
6K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K