# Homework Help: Solving logarithmic and exponential equations

1. Oct 15, 2013

### Mootjeuh

1. The problem statement, all variables and given/known data
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.

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

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.

2. Oct 15, 2013

### Staff: Mentor

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.
• 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.
• Where did this (above) come from?

Last edited: Oct 15, 2013
3. Oct 15, 2013

### Curious3141

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

Parentheses are important!

• OK so far.

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:

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

• 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.

4. Oct 15, 2013

### Mootjeuh

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?

5. Oct 15, 2013

### Staff: Mentor

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

6. Oct 15, 2013

### Mootjeuh

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. Oct 15, 2013

### Staff: Mentor

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. Oct 15, 2013

### Staff: Mentor

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.

9. Oct 15, 2013

### Mootjeuh

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!