Mathematics
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OpenStudy (anonymous):
what are the two values of x that satisfy log_x 2=log_3 X
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OpenStudy (anonymous):
can someone help me
OpenStudy (anonymous):
Use the definition of the log like we did with the others here.
OpenStudy (anonymous):
\[2 = log_3x \iff\ ?\]
OpenStudy (anonymous):
x=3^(log_x2)
idk....there are two xs and it's seems ambiguous....like, if i do 2=x^(log_3 X) then, it's just weird..
OpenStudy (anonymous):
I'm not sure there are 2 values for x that apply here though.
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OpenStudy (anonymous):
Wait, where's the 2 x's
OpenStudy (anonymous):
I only see one.
OpenStudy (anonymous):
log_x 2=log_3 X
OpenStudy (anonymous):
oh, that's different.
OpenStudy (anonymous):
But even so, the rule still applies.
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OpenStudy (anonymous):
\[Let\ k = log_3 x \implies 3^k = x\]
But that means that
\[k = log_x 2 \implies x^k = 2\]
OpenStudy (anonymous):
So if we take the first equation and raise it to the power of k we have
\[3^{kk} = x^k = 2\]
OpenStudy (anonymous):
i get your previous answer. But why are you raising them to the pwr k??
OpenStudy (anonymous):
So that I can set them equal.
OpenStudy (anonymous):
oh~
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OpenStudy (anonymous):
Because now we can take the ln of both side and have
\[k^2(ln 3) = (ln 2)\]
OpenStudy (anonymous):
so k = ?
OpenStudy (anonymous):
the sqroot of ln2/ln3. But how did you go from x^k=2 to ln2?
OpenStudy (anonymous):
I didn't. I went from \[3^{k^2} = 2 \implies k^2(ln\ 3) = (ln\ 2)\]
OpenStudy (anonymous):
oh, ok. got it.
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OpenStudy (anonymous):
Also \[k = \pm \sqrt{(\frac{ln 2}{ln 3})}\]
Which is where you get your two solutions from.
OpenStudy (anonymous):
Though you still have to solve for x.
OpenStudy (anonymous):
wait, so there's two values for x too?
OpenStudy (anonymous):
Yes, because there are 2 values for k, and k is related to x.
OpenStudy (anonymous):
\[k = log_3 x\] Remeber?
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OpenStudy (anonymous):
haha right~
OpenStudy (anonymous):
So \[x = 3^k\]
OpenStudy (anonymous):
So the two values that satisfy that equation for x are?
OpenStudy (anonymous):
3^sqroot ln3/ln2 and the other one is the same except it's the negative root of ln3/ln2. yes? (please say yes ^^:)
OpenStudy (anonymous):
It's ln2/ln3, but otherwise yes.
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OpenStudy (anonymous):
YAY <3