# Log sums

1. Jul 19, 2010

### 1/2"

1. The problem statement, all variables and given/known data
I have 2 problems ,which are:
1) Solve :log 7 (x 3+27) + log 7 (x+3)
2) Solve for x:log5 ( 5 1/x+ 125) log 5 6+1+1/2x

2. Relevant equations

3. The attempt at a solution
The first one tried like this :
log 7 { (x 3+27)/(x+3) }= 2
or 7 2 = (x 3+27)/(x+3)
in second one
log5{( 5 1/x+ 125) /6}=1+1/2x
but I can't proceed here too.
I need an urgent help!
Thank You!

2. Jul 19, 2010

### eumyang

Please check the original problems as I see typos abound. For instance, in the first problem,

log 7 (x 3+27) + log 7 (x+3)

...I see no right hand side. Also, if what you have above is correct, then you're next step...

log 7 { (x 3+27)/(x+3) }= 2

...is already wrong. (Is is supposed to be divide?)

Finally, if the last step above is supposed to have a divide, then review factoring special products -- specifically, the sum of two cubes.

69

3. Jul 19, 2010

### Mentallic

For the first one, you could multiply through by x+3 to give

$$49(x+3)=x^3+27$$

and then solve that cubic by simplifying and turning it into the form $$ax^3+bx^2+cx+d=0$$
but cubics are hard to solve, an easier way would be to realize that $$x^3+27$$ is a sum of two cubes, mainly $$x^3+3^3=(x+3)(x^2-3x+9)$$ which means you can cancel the x+3 from the numerator and denominator, giving you only a quadratic to solve, which is much simpler. Remember to look back at the original question to figure out what the domain of x is.

For the second, do the same thing as you did above, turn the log into powers such that

$$\frac{5^{\frac{1}{x}}+125}{6}=5^{1+\frac{1}{2x}}$$

Now simplify the right hand side. You should be able to factorize out 51/x and then solve.

4. Jul 20, 2010

### The Bob

I agree with eumyang. Is the question to simply the above expressions? Also, with the second problem, are you missing any brackets/parenthesis at all?

The Bob

5. Jul 21, 2010

### HallsofIvy

Staff Emeritus
At least in the second, 1/2" specifically says "solve for x". There are no equations to be solved!

6. Jul 21, 2010

### Mentallic

It's an honest mistake to mix up + and = considering they're the same key on the keyboard, but he did have lots of typos so it could have been re-interpreted as anything.

And for the first equation, let's just assume he started to solve the correct problem, i.e. f(x)=2.

7. Jul 23, 2010

### 1/2"

Thanks for the help but I figured it out!
Thanks everyone!