Hard logarithm question

[Baron]

If log_2n1994 = log_n(486sqrt(2)), compute n^6

My work:

As I can't figure out Latex, I'll apologize in advance for the messy work.

Using the definition of log:

n^(log 1994/log (2n)) = 486sqrt(2)

Log both sides

(log 1994/log (2n)) log n = log (486sqrt(2))

log 1994*log n = log (2n) * log (486 sqrt(2))

log 1994*log n = log (486 sqrt(2))*(log 2 + log n)

Let a = log n

a*log 1994 = log (486 sqrt(2))*(log 2 + a)

a log 1994 = log (486 sqrt(2))*log 2 + a log (486 sqrt(2))

a log 1994 - a log (486 sqrt(2)) = log 2*log (486 sqrt(2))

a (log 1994 - log (486 sqrt(2)) = log 2*log (486 sqrt(2))

a = {log 2*log (486 sqrt(2))}/(log 1994 - log (486 sqrt(2)))

log n ={log 2*log (486 sqrt(2))}/(log 1994 - log (486 sqrt(2)))

n = 10^[{log 2*log (486 sqrt(2))}/(log 1994 - log (486 sqrt(2)))]

Take both sides to the 6th power

n^6 = 10^[6{log 2*log (486 sqrt(2))}/(log 1994 - log (486 sqrt(2)))]

So that's my answer. Only problem is that I am suppose to simplify it as much as possible. I have no idea how to do that. Using a calculator (although the teacher said not to) the answer is
approximately 1.2*10^11. So I'm pretty sure the answer would involve exponents. I tried simplifying it but the fact there is multiplication between two logs and log 1994 which seems to only be simplified into log 997 + log 2 is making me go around in circles.

Any help would be appreciated in solving this. It's particularly frustrating because I feel as I am so close to solving this. And there has to be a reason they asked for n^6 instead of n so the answer would be a lot nicer.

 

[soroban]

Hello, Baron!

I approached it differently ... and also got into a mess.

\(\displaystyle \text{If }\,\log_{2n}\!1994 \:=\:\log_n\!486\sqrt{2},\,\text{ compute }n^6.\)

One form of the Base-change Formula: .\(\displaystyle \log_ba \:=\:\dfrac{\ln a}{\ln b}\)

\(\displaystyle \text{We have: }\:\log_{2n}\!1994 \:=\:\log_n\!486\sqrt{2}\)


. . \(\displaystyle \dfrac{\ln1994}{\ln2n} \:=\:\dfrac{\ln486\sqrt{2}}{\ln n} \)


. . \(\displaystyle \ln1994\!\cdot\!\ln n \:=\:\ln2n\!\cdot\!\ln 486\sqrt{2}\)

. . \(\displaystyle \ln1994\!\cdot\!\ln n \:=\:\ln2n \bigg[\ln486 + \frac{1}{2}\ln2\bigg]\)

. . \(\displaystyle \ln1994\!\cdot\!\ln n \:=\:\ln486\!\cdot\!\ln2n + \frac{1}{2}\ln2\!\cdot\!\ln2n \)

. . \(\displaystyle \ln1994\!\cdot\!\ln n \:=\:\ln486(\ln 2 + \ln n) + \frac{1}{2}\ln2(\ln 2 + \ln n) \)

. . \(\displaystyle \ln1994\!\cdot\!\ln n \:=\:\ln486\!\cdot\!\ln2 + \ln486\!\cdot\!\ln n + \frac{1}{2}(\ln 2)^2 + \frac{1}{2}\ln2\!\cdot\!\ln n\)


. . \(\displaystyle \ln1994\!\cdot\!\ln n - \ln486\!\cdot\!\ln n - \frac{1}{2}\ln 2\!\cdot\!\ln n \;=\;\ln2\!\cdot\!\ln486 + \frac{1}{2}(\ln 2)^2\)

. . [tx]. . .\(\displaystyle \bigg[\ln1994 - \ln486 - \frac{1}{2}\ln2\bigg]\,\ln n \;=\; \ln2\bigg[\ln486 + \frac{1}{2}\ln2\bigg]\)

. . . . . . . . . . . . . . . . . .\(\displaystyle \ln\left(\dfrac{1994}{486\sqrt{2}}\right)\ln n \;=\;\ln2\!\cdot\!\ln486\sqrt{2}\)

. - - . . . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle \ln n \;=\;\dfrac{\ln2\!\cdot\!\ln486\sqrt{2}}{\ln\frac{1994}{486\sqrt{2}}} \)

\(\displaystyle \text}Let }\,c \,=\,\dfrac{\ln2\!\cdot\!\ln486\sqrt{2}} {\ln\frac{1994}{486\sqrt{2}}} \)

Then we have: .\(\displaystyle \ln n \:=\:c \quad\Rightarrow\quad n \:=\:e^c\)

Therefore: .\(\displaystyle n^6 \;=\;e^{6c}\)