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lmsavk
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#1
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#1
Given\ that \ t=\log_3x,\ find \ expressions \ in\ terms\ of\ t\ for\  x.
(i)\ log_3x^2 = \frac{t}{2}, \ gg\ ez\ br0
(ii)\ log_9x = ???,\ what\ do\ I\ do\ here?
Why\ does\ \frac{(log_3x)}{(log_39)}\ = log_9x?
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TenOfThem
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#2
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(Original post by lmsavk)
Given\ that \ t=\log_3x,\ find \ expressions \ in\ terms\ of\ t\ for\  x.
(i)\ log_3x^2 = \frac{t}{2}, \ gg\ ez\ br0
(ii)\ log_9x = ???,\ what\ do\ I\ do\ here?
Why\ does\ \frac{(log_3x)}{(log_39)}\ = log_9x?
Do not put the writing inside the latex it is horrible

t = \log _3x

(ii) use the change of base rule - you should know what \log _3 9 is
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Smaug123
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(Original post by lmsavk)
Given\ that \ t=\log_3x,\ find \ expressions \ in\ terms\ of\ t\ for\  x.
(i)\ log_3x^2 = \frac{t}{2}, \ gg\ ez\ br0
(ii)\ log_9x = ???,\ what\ do\ I\ do\ here?
Why\ does\ \frac{(log_3x)}{(log_39)}\ = log_9x?
Um, you can write text in non-LaTeX as well, you know :P
\dfrac{\log_3(x)}{\log_3(9)} is a specific instance of the general rule:
\dfrac{\text{lg}(x)}{\text{lg}(k  )} = \log_k(x), where I use \text{lg} to mean "log to any base".
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lmsavk
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(Original post by TenOfThem)
t = \log _3x

(ii) use the change of base rule - you should know what \log _3 9 is
\log _3 9=2 but I don't understand why that changes the base?
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ThatPerson
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#5
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(Original post by lmsavk)
Given\ that \ t=\log_3x,\ find \ expressions \ in\ terms\ of\ t\ for\  x.
(i)\ log_3x^2 = \frac{t}{2}, \ gg\ ez\ br0
(ii)\ log_9x = ???,\ what\ do\ I\ do\ here?
Why\ does\ \frac{(log_3x)}{(log_39)}\ = log_9x?
Take a look at this video:

https://www.khanacademy.org/math/alg...-formula-proof
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