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# I need help, if you could help i would greatly appreciate it

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what is the value of n such that, 81^n=9^27 * 27^9?

Dec 30, 2022

#1
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The correct value of n is 16/3.

Dec 30, 2022
#2
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thx but i need solution is okay if you provide me with a solution?

Guest Dec 30, 2022
#3
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Why are you having problems with this simple question? If your calculator display isn't large enough, then use the calculator on this site!

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

6 - Therefore: n = 20.25.  And that is it!

Dec 30, 2022
#4
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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.

Dec 30, 2022