Answer me PLEASE !!!!!?

Question

With full correct and simple steps find n, such that ;

n < log of 96 to the base (3) < n + 1

Answer 1

log(base3) of 96 has to be between 4 and 5 because it represents the power you have to raise 3 to to get 96.

Since 3^4 = 81 and 3^5 = 243, the power you raise 3 to to get 96 must be between those.

n = 4
n + 1 = 5

Answer 2

In the log 3 system, rather than the place values being the familiar 1s, 10s, 100s, 1000s, etc., they are 1s, 3s, 9s, 27s and 81s, etc.

See how that works? In base 10, the place values are 10 to the zero power (anything to the zero power is always "1"), 10 to the first power (10), ten to the second power (10 x 10 = 100), etc.

Similarly in base 3, the place values are 3 to the zero power (remember, that's 1!), 3 to the first (3), 3 to the 2nd (9), 3 to the third (27) and 3 to the fourth power (3 x 3 x 3 x 3 = 81).

To convert 96 from base 10 to base 3, we first notice that the largest base 3 place value in the number 96 is "81", once we deal with 81 of the original 96 that we're trying to convert, we're left with a remainder of 15 (96-81 = 15) that we still need to work with.

So we start out writing a 1 (because there's one "81") in the fifth position.

1 _ _ _ _

Now let's deal with the 15 remainder. We see that there are no "27s" in the 15 that remain, so we put a zero in the 27's place.

1 0 _ _ _

There is, however, one "9" in our remaining 15, so we'll put a 1 in the 9's place:

1 0 1 _ _

After dealing with 9 of the original remainder of 15, we still have 6 left (15-9 = 6), so we move down to the 3's place - there are two 3s in six.

1 0 1 2 _

There are no 1's left to worry about, so we'll write a zero in the final postion - we've already accounted for all of the original 96, with no remainder.

Result: 10120 is the base 3 equivalent of 96 in base 10.

After that, you've lost me, unless n doesn't need to be a whole number. If n is 10119.5, for example, then this equation would be correct:

10119.5 < 10120 < 10120.5

Hope that makes sense! good luck!

(then again, maybe this isn't what you're asking at all)

Answer 3

Hi,
Here's a simple way to do it.

n < log base 3 of 96 < n+1

3^n < 3^(logbase3 of 96) < 3^(n+1)

there's a property of logs that says if you raise any number, say Y to the logbaseYof X power you get X, so your inequality now becomes

3^n < 96 < 3^(n+1)

3^4 < 96 < 3^(4+1)

so n = 4

Answer 4

So what power of 3 gives 96?

3^2=9 so we need to go bigger
3^3=27
3^3=81
3^4=243

We skipped over 96. So that means the power of 3 that will give us 96 is somewhere between 3 and 4:
3 < log of 96 to the base (3) < 4

So that means n=3

Answer 5

n < log(3) 96 < n + 1
considering,
3^n < 96
3^n < 3 * 32
3 ^ n -1 < 32

considering,
log(3)96 < n +1
3 * 32 < 3 ^ n + 1
32 < 3 ^ n

therefore,
3 ^ n -1 < 32 < 3 ^ n
if n is an integer, than n = 4
because 3 ^ 4 = 81, which is greater than 32 and 3 ^ 3 = 27, which is smaller than 32.
There may be a graphical way of showing it and if n is not an integer, I am sorry it has many possible answers, I have just finished my O levels, I am probably not qualified enough to answer this

Answer 6

Start with some definitions to make this look cleaner
Log.a (x) == what you're calling "log to the base "a" "
Ln (x) == the natural log (log base e, where e is Euler's number).

Start by writing the middle as:
Ln(96) / Ln(3) ; that is a basic property of logs
multiply through the inequality by Ln(3)

n*Ln(3) < Ln(96) < (n+1)*Ln(3)

Now look at e raised to each of these guys

e^[n*Ln(3)] < e^[Ln(96)] < e^[(n+1)*Ln(3)]

which is the same as

3^n < 96 < 3^(n+1)

You can figure it out from there, it really isn't that hard, just look at a couple of choices for n.

I don't mind showing people how to get an answer, but I don't like straight up giving it to them. Good Luck!

Answer 7

n=4

Answer 8

I failed in logrithems

Answer 9

!!!!!!!!!!!!!!!!!!!

Answer 10

i do not know