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

[Teh]

Convert the Following expression to the indicated base, using base 14 for a > 0 & a \ne 1.$${a}^{\frac{5}{log}_{9{}^{a}}}$$

 

[Greg]

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?

 

[Teh]

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]

 

[Greg]

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"?

 

[Teh]

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}

 

[I like Serena]

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.