MHB Changing Base with Logs: How to Convert Expressions Using Base 14?

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The discussion focuses on converting the expression \( a^{\frac{5}{\log_9 a}} \) to base 14. Participants clarify that the goal is to convert the logarithm from base 9 to base 14, using the relationship \( \log_b x = \frac{\ln x}{\ln b} \). Through substitution, they derive that \( a^{\frac{5}{\log_9 a}} \) simplifies to \( 9^5 \), demonstrating that the base conversion does not affect the final result. The conversation highlights the flexibility of using different logarithmic bases, confirming that the outcome remains consistent regardless of the base used. Overall, the key takeaway is the successful transformation of the expression without needing to explicitly convert to base 14.
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Convert the Following expression to the indicated base, using base 14 for a > 0 & a \ne 1.$${a}^{\frac{5}{log}_{9{}^{a}}}$$
 
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Hi Teh. I've shortened up the title of your thread.

Did you intend

$a^{5/\log_9a}$

?

Why did you post this in trigonometry?
 
greg1313 said:
Hi Teh. I've shortened up the title of your thread.

Did you intend

$a^{5/\log_9a}$

?

Why did you post this in trigonometry?
THIS IS WHAT I MEANT! THANKS! though it was trig problem because in class my professor was going over trig...sorry if it was not [/QUOTE]
 
I'm still not clear on what's intended. Are we to convert the base 9 log to a base 14 log? If not, what is the "indicated base"?
 
greg1313 said:
I'm still not clear on what's intended. Are we to convert the base 9 log to a base 14 log? If not, what is the "indicated base"?

same also I don't know what is is asking for I ask my professor all he gave me was log{}_{b}{x}^{} = \frac{lnx}{lnb}
 
Teh said:
same also I don't know what is is asking for I ask my professor all he gave me was $\log_{b}{x} = \frac{\ln x}{\ln b}$

Okay, so let's substitute that:
$$a^{\frac 5 {\log_9 a}} = a^{\frac 5 {\frac{\ln a}{\ln 9}}}
=a^{\frac {5\ln 9} {\ln a}}
$$
Now let's take the logarithm of all of that and see where it brings us:
$$\ln \left(a^{\frac 5 {\log_9 a}}\right)
=\ln\left(a^{\frac {5\ln 9} {\ln a}}\right)
=\frac {5\ln 9} {\ln a} \ln(a)
= 5\ln 9
= \ln \left(9^5\right)
$$

Hey! That means that:
$$a^{\frac 5 {\log_9 a}} = 9^5$$
And that's even without referring to $\log_{14}$. (Cool)
Note that we could have used $\log_{14}$ everywhere instead of $\ln$. The result is the same.
 
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