Solving logarithmic and exponential equations

[Mootjeuh]

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: 2^x + 2^x+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:

• 2^x + 2^x+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 2^y 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.

 

[Mark44]

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: 2^x + 2^x+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:

• 2^x + 2^x+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 2^y 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.

 

[Curious3141]

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: 2^x + 2^x+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:
  

  • 2^x + 2^x+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.

 

[Mootjeuh]

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 2^x go?

 

[Mark44]

Thanks, I didn't think of using an antilog, solved it smoothly.

I kind of don't follow what happened here, where did 2^x go?

It's still there.
2^{x + 1} = 2 * 2^x

 

[Mootjeuh]

Yeah but I mean the original formula was 2^x + 2^x+1 = 3, you moved it to the other side of the equals sign and it became 2 times 2^x. Why not 3-2^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

 

[Chestermiller]

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

 

[Mark44]

Yeah but I mean the original formula was 2^x + 2^x+1 = 3, you moved it to the other side of the equals sign and it became 2 times 2^x. Why not 3-2^x?

Because doing that won't help you. If you subtract 2^x from both sides, you get
2^{x + 1} = 3 - 2^x
This isn't wrong, but it's not helpful, either, in getting you closer to a solution.

This is what you need to do:
2^x + 2^x+1 = 3
==>2^x + 2*2^x = 3
==> 2^x(1 + 2) = 3
==>2^x * 3 = 3
==>2^x = 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

 

[Mootjeuh]

Aaaah right 2^x+1 = 2*2^x is the same as 2¹*2^x which makes the rest distributable now

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