Hi i really need help with dealing with logs

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akj
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Main Question or Discussion Point

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!!!!!
 

Answers and Replies

HallsofIvy
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Since logarithms have nothing to do with "Linear and Abstract Algebra", I am moving this to "general math"
 
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What do the ------ parts mean?
 
quasar987
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it's an attempt at writing division. For instance,

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

should read

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

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.
 
Last edited by a moderator:
akj
4
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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
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I asked before, "Could you solve A- (1/2)A= 2?" Are you saying you cannot solve that equation?
 

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