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81^n=9^27 * 27^9?
1 - With a calculator, calculate the right-hand side first: 9^27 * 27^9 =443426488 2430377699 4824963061 9149892803
2 - Take the log (base 10) of the above number: log(443426488 2430377699 4824963061 9149892803)
3 - Log = 38.6468216322 (you only need the log to 10 decimal places).
4 - Take the log of 81 to 10 decimal places: log(81) = 1.9084850189
5 - Divide the log in (3) above by the log in (4) above: 38.6468216322 / 1.9084850189=20.25
6 - Therefore: n = 20.25. And that is it!
+105 I'm thinking to solve it a different way...
We have:
\(81^n=9^{27} \cdot 27^9\)
We know 81 is 3^4
\((3^4)^n = 9^{27} \cdot 27^9\)
So the left side simplifies to
\(3^{4n} = 9^{27} \cdot 27^9\)
We know 9 is 3^2 and 27 is 3^3 so we use that information into our equation:
\(3^{4n} = (3^2)^{27} \cdot (3^3)^9\)
So the right side simplifies to:
\(3^{4n} = (3^{54}) \cdot (3^{27})\)
Then, we know that if we have the same base when multiplying, we add the exponents so:
\(3^{4n} = (3^{81})\)
So:
\(4n = 81 \)
Divide by 4 to both sides:
\(n = 81/4 = 20.25\)
This is (I believe) the standard and faster way to solve it in a real competition.
