Hi i really with dealing with logs

[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

but i am really stuck with dealing with the x please help!

 

[HallsofIvy]

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

 

[Couperin]

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

 

[quasar987]

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

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

should read

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

 

[HallsofIvy]

Trying to interpret:

Hi i really need help with dealing with logs:

log x - log x = 2
3 9

log₃ x- log₉ x= 2

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

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}

Now i have

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

log₃(x)- (1/2)log₃(x)= 2

but i am really stuck with dealing with the x please help!

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.

 

[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?

 

[HallsofIvy]

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