## louis413 3^5n=3^5(3^2n)^2 solve for n one year ago one year ago

1. Wired

Please rewrite the question using the equation tool, as it's tough to tell if you mean: $3^{5n}$ or $3^{5}n$ for example. Same thing goes for the other side of the equation.

2. louis413

its the first one 5n is on three

3. KingGeorge

$3^{5n}=3^5\cdot(3^{2n})^2$Correct?

4. louis413

$3^{5n}=3^{5}(3^{2n})^{2} solve for n$

5. louis413

yes

6. KingGeorge

First, you need to simplify the expression on the right side. Recall that $(a^b)^c=a^{bc}$This means that you can rewrite the right hand side as $3^5\cdot3^{4n}$

7. KingGeorge

Now we also have the rule that $a^b\cdot a^c=a^{b+c}$That means we can rewrite the RHS again. This time as $3^{5+4n}.$That means we have the relation $3^{5n}=3^{5+4n}$

8. KingGeorge

Since 3 is the base on both sides, we can just get rid of it (otherwise known as taking the log base 3 of both sides). Hence, we only have to solve the equation $5n=5+4n$Can you do this yourself?

9. louis413

wait im confused like if its (3xy)^3 then the "3" would be 27 but there the 3 is still "3" whys that?

10. KingGeorge

In the case of $$(3xy)^3$$, you can think of it as $(3xy)(3xy)(3xy)=(3\cdot3\cdot3)\cdot(x\cdot x\cdot x)\cdot(y\cdot y\cdot y)=3^3\cdot x^3\cdot y^3=27x^3y^3$

11. louis413

oh hey can u help me on this prob? $3*9^{2n}=(3^{n+1})^{3}$

12. KingGeorge

First off, write $3\cdot9^{2n} =3\cdot(3^2)^{2n}$Can you show the next step in simplifying this?

13. louis413

ok um its $(3^{2n+1}) on the \right side im confused for the \left side$

14. Wired

It may help you if you temporarily replace the exponents with simpler variables. For example, you could say 2n = X, and n+1 = Y. That way you can rewrite that equation as: $3*9^{x} = (3^{y})^{3}$ From there you can use the rules George posted.

15. louis413

*3^2n+2

16. KingGeorge

Wait, on the right side, you have$\Large (3^{n+1})^3$What you gave, was a simplification for $\Large (3^{n+1})^2$

17. louis413

oh yeah my bad yeah its 3

18. louis413

please i get the right side is 3^3n+3 but what i do to the rigleft side?

19. louis413

left side

20. KingGeorge

For the left side, we have $\large 3\cdot(3^2)^{2n}$Using the same principles you just used for the right side, what is $\large (3^2)^{2n}?$

21. louis413

its 3^4n

22. KingGeorge

Perfect. Now what is $\large 3\cdot3^{4n}?$

23. louis413

3^4n+1?

24. KingGeorge

Right again. That means we can take log base 3 of both sides, and get $4n+1=3n+3$From here, it's just a simple solving of a linear equation.

25. Wired

FYI if you do substitute 2n = X, and n+1 = Y: $3*9^{x} = 3^{3y}$ Since $9 = 3^{2}$ $3*(3^{2})^{x} = 3^{3y}$ $3*3^{2x} = 3^{3y}$ $3^{1}*3^{2x} = 3^{3y}$ $3^{2x+1} = 3^{3y}$ Take the log base 3 of both sides to get: 2x+1 = 3y Replace x and y with their values to get: 2(2n)+1 = 3(n+1) 4n+1 = 3n+3

26. louis413

oh thank you very much hey one more

27. louis413

quick question

28. louis413

why is 3 in (3xy)^3 is 27... but 3in (3^2n)^2 is still 3 ?

29. Wired

You may want to start a new question for each question to make it easier to read for others later on.

30. Wired

Is that $(3^{2n})^{2}$ ?

31. louis413

yea

32. louis413

is it because we treat 3 as an x? so 3 doesnt change?

33. louis413

and also usually the 3 has a degree of 0 but here it is one

34. Wired

$3^{4n}$ Because you don't know what the full exponent is, it's easier to keep it as it is. You could rewrite it as $(3^{4})^{n}$ and then solve for 3^4, so you'd get: $81^{n}$

35. Wired

3 has a degree of 0, huh?

36. louis413

in the degree of a monomial the constant has a degree of 0 right?

37. KingGeorge

It's not the constant that has a degree, it's the variable that has degree 0. Because the degree is 0, the variable can be replaced with $$x^0=1$$, so you see a constant term.