If log(base 2n)(1944)=log(base n)(486sqr(2), determine the value of n^6

Question

How do get an answer of approx 2.23*10^11?

Answer 1

Use the change of base formula to convert both sides of the given equation to base 10 logs:

 

We get (log 1944)/(log(2n) = (log (486√2))/(logn).

 

So, (log 1944)(logn) = (log2 + logn)(log(486√2))

 

(log1944)(logn) = (log2)(log(486√2)) + (logn)(log(486√2))

 

(logn)(log1944 - log(486√2)) = (log2)(log(486√2)

 

logn = (log2)(log(486√2))/(log(1944) - log(486√2))

 

logn = 1.891434178, so n = 10^1.891434178

 

Therefore, n⁶ = 10^11.34860507 ≈ 2.23 ⋅10¹¹