# Hi i really need help with dealing with logs

1. Feb 12, 2007

### akj

Hi i really need help with dealing with logs:

log x - log x = 2
3 9

so i change the base, since 9 is a power of 3.

y = log_9(x)

Then

9^y = x

(3^2)^y = x

3^(2y) = x

2y = log_3(x)

log_3(x)
y = --------
2

Now i have

log_3(x)
log_3(x) - -------- = 2
2

2. Feb 12, 2007

### HallsofIvy

Staff Emeritus
Since logarithms have nothing to do with "Linear and Abstract Algebra", I am moving this to "general math"

3. Feb 12, 2007

### Couperin

What do the ------ parts mean?

4. Feb 12, 2007

### quasar987

it's an attempt at writing division. For instance,

log_3(x)
y = --------
2

$$y=\frac{\log_3{x}}{2}$$

5. Feb 12, 2007

### HallsofIvy

Staff Emeritus
Trying to interpret:
log3 x- log9 x= 2

$$y= \frac{log_3(x)}{2}$$
There is a general formula that
$$log_a(x)= \frac{\log_b(x)}{log_b(a)}$$
where b is any positive number. In particular, taking b= 3 here
$$log_9(x)= \frac{\log_3(x)}{log_3(9)}= \frac{\log_3(x)}{2}$$

log3(x)- (1/2)log3(x)= 2

Could you solve A- (1/2)A= 2? If so then set log3(x) equal to that and solve for x by taking 3 to the power of each side.

Last edited: Feb 12, 2007
6. Feb 14, 2007

### akj

hi,

i have calculated that x = 81 which is correct, but i solved this with trial & error. How would i have solved the final section of this equation using algebra?

7. Feb 14, 2007

### HallsofIvy

Staff Emeritus
I asked before, "Could you solve A- (1/2)A= 2?" Are you saying you cannot solve that equation?