# Homework Help: Logarithmic Equation

1. Nov 14, 2008

### nikita33

Just wondering if someone can tell me if I got this correct. It was a problem on my midterm a month ago and im going over the probs I got wrong, and I dont know the correct answers.
1. The problem statement, all variables and given/known data

4(x+1)= 8(x-1)

2. Relevant equations

3. The attempt at a solution
log 4(x+1) = log 8(x-1)

(x+1) log 4 = (x-1) log 8

x log 4 + log 4 = x log 8 - log 8 (sub/add to get x on one side)

log 8 + log 4 = x log 8 - x log 4

log 8 + log 4 = x(log 8 - log 4) (divide to get x)

log 8 + log 4 = X
---------------
log 8 - log 4

x= 5

2. Nov 14, 2008

### Dick

I guess that works, though you didn't really say how you changed your final log expression into 5. But you can make your life a bit easier by picking a convenient base for the logarithms. In this case I'd pick log_2 (log base 2) and use that. So your second line just turns into 2(x+1)=3(x-1).

3. Nov 14, 2008

### nikita33

^
well, that would make it easier, now wouldnt it? prob why i got it wrong on the exam. i totally forgot about that.

To get 5, I divided
(log 8 + log 4)/(log 8 - log 4)
which is
1.505149978/.3010299957 = 4.999999998

thank you though for reminding me about the bases. I couldnt do it the way I did it if it was on the non calculator portion of the exam, so Im going to have to remember that. I have another log question, but its a diff problem, so ill make another post.

4. Nov 14, 2008

### Dick

You could also do it this way. (log(8)+log(4))=log(8*4)=log(32)=log(2^5)=5*log(2) and (log(8)-log(4))=log(8/4)=log(2). So you can do it without a calculator, though it pays to get rid of the logs early.

5. Nov 14, 2008

### nikita33

ah^
that is like the other question i posted. I forgot on the exam that addition is multiplication. Well, not on the expanding log questions but on solve for x questions.

6. Nov 14, 2008

### symbolipoint

Use of logarithms is not necessary for that example.
(2^2)^(x+1) Left Hand Side.
(2^3)^(x-1) Right Hand Side.

Skipping one step,
2^(2x+2) = 2^(3x-3)
Left and Right sides having the same base, the exponents are equal.

7. Nov 14, 2008

### Dick

There's more than one way to skin a cat.

8. Nov 15, 2008

### CompuChip

Yep, here's yet another one;
4^(x+1) = 8^(x-1)
16 * 4^(x - 1) = 8^(x - 1)
16 = (8/4)^(x - 1) = 2^(x - 1)
x - 1 = 4

Of course, most methods mentioned are not general, because they rely on the convenient fact that 4= 2^2 and 8 = 2^3.