1. ## Logarithms...again

log(base 2) (x+4) + log(base 2) (x+8)=5

Here's my work:
log(2) (x+4)(x+8)=10
x^2+12x+22=0
(x+11) (x+2)

But my teacher said you don't use negative numbers, so what's my answer?

And, I don't know how to do this problem:

log(base 5)(2x+4) - log(base 5)(x-1)=1

My work:
log5 2(x+2)/(x-1)=5

Then what?

2. Originally Posted by beetz
log(base 2) (x+4) + log(base 2) (x+8)=5
$\log_2(x+4)+\log_2(x+8)=5\implies\log_2\Big[(x+4)(x+8)\Big]=5.$

So, $(x+4)(x+8)=32\implies x^2+12x=0\,\therefore\,x_1=0\,\wedge\,x_2=-12.$

We require that $x+4>0$ & $x+8>0$ such that the logarithms be well defined.

3. Originally Posted by Krizalid
$\log_2(x+4)+\log_2(x+8)=5\implies\log_2\Big[(x+4)(x+8)\Big]=5.$

So, $(x+4)(x+8)=32\implies x^2+12x=0\,\therefore\,x_1=0\,\wedge\,x_2=-12.$

We require that $x+4>0$ & $x+8>0$ such that the logarithms be well defined.
my teacher said what you set the equation equal to is the base, in this case 2, multiplied by the answer given, 5.

4. Originally Posted by beetz
my teacher said what you set the equation equal to is the base, in this case 2, multiplied by the answer given, 5.
no. if $\log_a b = c$ then $a^c = b$

in other words, what you set a log equal to is the power of the base (that is, the power you must raise the base to to get what's being logged). (you should pay attention in class)

5. I was paying attention and I promise you that is exactly what he said. I even have a problem he did on the board and he multiplied the log base and the answer given. In fact, he never did a formal lecture, just skimmed over the homework and did a few problems.

I think I get it...
Thanks.

6. Originally Posted by beetz
I was paying attention and I promise you that is exactly what he said. I even have a problem he did on the board and he multiplied the log base and the answer given. In fact, he never did a formal lecture, just skimmed over the homework and did a few problems.
ok, ok, if you say so. i believe you

I think I get it...
Thanks.
show us the solution