• Support PF! Buy your school textbooks, materials and every day products Here!

Solving logarithmic and exponential equations

  • Thread starter Mootjeuh
  • Start date
  • #1
5
0

Homework Statement


I'm trying to solve trying logarithmic and exponential equations but for the life of me I can't seem to figure out what I'm doing wrong.


Homework Equations


Log equation: log3(x+2)+1 = log3x²+4x
Exp equation: 2x + 2x+1 = 3


The Attempt at a Solution


For the first one, here's what I did:
  • 1 = log3(x²+4x) - log3(x+2)
  • log3((x²+4x)/(x+2)) = 1
  • log3((x² +4x)/(x+2)) = 1
  • log3((x+4x)/2) = 1 --Can I do this? Dividing the denominator and numerator by the same value within a log
  • log3(x+2x) = 1
  • log3(3x) = 1
  • log33 + log3x = 1
  • 1 + log3x = 1
  • log3x = 0
  • x = 1

And then for the second one:
  • 2x + 2x+1 = 3
  • I figured I'd put them all on the same base so I tried to find 2 to what power equals 3
  • So I did ln(3)/ln(2) = 1.584962501, which for simplicity's sake I'll replace by y here
  • And when I put in 2y in my calculator it does give me 3 so it must be correct
  • x + x + 1 = y
  • 2x + 1 = y
  • x = (y-1)/2
But when I put the result back in the original equations, none of the answers I found seem to be correct.
I'm at my wit's end here, I don't know what else to try.

Any help is greatly appreciated.
 

Answers and Replies

  • #2
33,484
5,174

Homework Statement


I'm trying to solve trying logarithmic and exponential equations but for the life of me I can't seem to figure out what I'm doing wrong.


Homework Equations


Log equation: log3(x+2)+1 = log3x²+4x
For the log expression on the right side, do you mean log3(x2 + 4x)? That seems to be what you're doing in your work below.
Exp equation: 2x + 2x+1 = 3


The Attempt at a Solution


For the first one, here's what I did:
  • 1 = log3(x²+4x) - log3(x+2)
  • log3((x²+4x)/(x+2)) = 1
  • log3((x² +4x)/(x+2)) = 1
  • log3((x+4x)/2) = 1 --Can I do this? Dividing the denominator and numerator by the same value within a log
  • I have no idea what you did. How did you go from (x2 + 4x)/(x + 2) to (x + 4x)/2? That's not a valid step.
    [*]log3(x+2x) = 1
    [*]log3(3x) = 1
    [*]log33 + log3x = 1
    [*]1 + log3x = 1
    [*]log3x = 0
    [*]x = 1
And then for the second one:
  • 2x + 2x+1 = 3
  • I figured I'd put them all on the same base so I tried to find 2 to what power equals 3
  • So I did ln(3)/ln(2) = 1.584962501, which for simplicity's sake I'll replace by y here
  • Where did this (above) come from?
    [*]And when I put in 2y in my calculator it does give me 3 so it must be correct
    [*]x + x + 1 = y
    [*]2x + 1 = y
    [*]x = (y-1)/2
But when I put the result back in the original equations, none of the answers I found seem to be correct.
I'm at my wit's end here, I don't know what else to try.

Any help is greatly appreciated.
 
Last edited:
  • #3
Curious3141
Homework Helper
2,843
87

Homework Statement


I'm trying to solve trying logarithmic and exponential equations but for the life of me I can't seem to figure out what I'm doing wrong.


Homework Equations


Log equation: log3(x+2)+1 = log3x²+4x
That should be written: log3(x+2)+1 = log3(x²+4x)

Parentheses are important!

Exp equation: 2x + 2x+1 = 3


The Attempt at a Solution


For the first one, here's what I did:
  • 1 = log3(x²+4x) - log3(x+2)
  • log3((x²+4x)/(x+2)) = 1
  • log3((x² +4x)/(x+2)) = 1


  • OK so far.

    [*]log3((x+4x)/2) = 1 --Can I do this? Dividing the denominator and numerator by the same value within a log
    You can divide a rational expression within a log argument. But your algebra is totally wrong here, because ##\frac{x^2+4x}{x+2} \neq \frac{x+4x}{2}##. Please review your algebra.

    Instead of this, I would just take antilogs of both sides to get:

    ##\frac{x^2+4x}{x+2} = 3##

    and solve that quadratic. Be careful about which solution(s) is/are admissible.

    And then for the second one:
    • 2x + 2x+1 = 3
    • I figured I'd put them all on the same base so I tried to find 2 to what power equals 3


    • Doesn't help because ##a^x + a^y = a^z## does NOT imply that ##x + y = z##. In fact, there's no "easy" relationship between x, y and z here.

      To solve, observe that ##2^{x+1} = 2.2^x## (the dot signifies multiplication). Now put ##2^x = y## to get a simple linear equation.
 
  • Like
Likes 1 person
  • #4
5
0
Thanks, I didn't think of using an antilog, solved it smoothly.

To solve, observe that ##2^{x+1} = 2.2^x## (the dot signifies multiplication). Now put ##2^x = y## to get a simple linear equation.
I kind of don't follow what happened here, where did 2x go?
 
  • #5
33,484
5,174
Thanks, I didn't think of using an antilog, solved it smoothly.

I kind of don't follow what happened here, where did 2x go?
It's still there.
2x + 1 = 2 * 2x
 
  • #6
5
0
Yeah but I mean the original formula was 2x + 2x+1 = 3, you moved it to the other side of the equals sign and it became 2 times 2x. Why not 3-2x?

I'm sure it's obvious and I completely missed it but it's 1am here and trying to finish this before going to sleep
 
  • #7
20,116
4,203
The quantity log3((x² +4x)/(x+2)) represents the power to which you have to raise the number 3 to get ((x² +4x)/(x+2)). The equation log3((x² +4x)/(x+2))=1 says that the number 1 is the power to which you have to raise the number 3 to get ((x² +4x)/(x+2)). So, 31=((x² +4x)/(x+2))

Chet
 
  • #8
33,484
5,174
Yeah but I mean the original formula was 2x + 2x+1 = 3, you moved it to the other side of the equals sign and it became 2 times 2x. Why not 3-2x?
Because doing that won't help you. If you subtract 2x from both sides, you get
2x + 1 = 3 - 2x
This isn't wrong, but it's not helpful, either, in getting you closer to a solution.

This is what you need to do:
2x + 2x+1 = 3
==>2x + 2*2x = 3
==> 2x(1 + 2) = 3
==>2x * 3 = 3
==>2x = 1
Now you can solve for x.
I'm sure it's obvious and I completely missed it but it's 1am here and trying to finish this before going to sleep
 
  • Like
Likes 1 person
  • #9
5
0
Aaaah right 2x+1 = 2*2x is the same as 21*2x which makes the rest distributable now

Great thanks I can finally go catch some Zs now, you've been a real help!
 

Related Threads on Solving logarithmic and exponential equations

Replies
5
Views
1K
Replies
12
Views
1K
  • Last Post
Replies
3
Views
805
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
6
Views
2K
Replies
7
Views
23K
Replies
3
Views
648
  • Last Post
Replies
6
Views
663
Top