Is my equality statement with logarithms as simplified as it can be?

vaderboi

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Say I was trying to isolate N and here is my equation:

D <= 3^(2^(N-1))

Is this as simplified as the equation can get?:

((log * (log D / log 3)) + log 2) / log 2 <= N

Here is my logic:

log D <= log 3^(2^(N-1))

log D <= 2^(N-1) * log 3

log D / log 3 <= 2^(N-1)

log * (log D / log 3) <= (N-1)*(log 2)

log * (log D / log 3) <= (N*log2) - log 2

(log * (log D / log 3)) + log 2 <= N*log2

((log * (log D / log 3)) + log 2) / log2 <= N
 
Say I was trying to isolate N and here is my equation:

D <= 3^(2^(N-1))

Is this as simplified as the equation can get?:

((log * (log D / log 3)) + log 2) / log 2 <= N

Here is my logic:

log D <= log 3^(2^(N-1))

log D <= 2^(N-1) * log 3

log D / log 3 <= 2^(N-1)

log * (log D / log 3) <= (N-1)*(log 2)

log * (log D / log 3) <= (N*log2) - log 2

(log * (log D / log 3)) + log 2 <= N*log2

((log * (log D / log 3)) + log 2) / log2 <= N

"log" isn't a number you can multiply; it's a function. So at the least, we have remove all those meaningless multiplications. (The fact that it isn't a multiplication is the reason you can't eliminate a log by dividing.)

Also, this isn't an equation or "equality statement"; it's an inequality (or, as some call it, an inequation).

Apart from that, your work is acceptable. It would look simpler, though, if you use the base 3 log, then the base 2 log, rather than using an unspecified log throughout. The result then looks like N >= log2[log3(D)] + 1.
 
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Say I was trying to isolate N and here is my equation:

D <= 3^(2^(N-1))

Is this as simplified as the equation can get?:

((log * (log D / log 3)) + log 2) / log 2 <= N

Here is my logic:

log D <= log 3^(2^(N-1))

log D <= 2^(N-1) * log 3

log D / log 3 <= 2^(N-1)

log * (log D / log 3) <= (N-1)*(log 2)

log * (log D / log 3) <= (N*log2) - log 2

(log * (log D / log 3)) + log 2 <= N*log2

((log * (log D / log 3)) + log 2) / log2 <= N
Did you even read Dr. Peterson's or my posts in your other thread?

There is a family of logarithm functions, not one logarithm function. There are in fact an infinite number of logarithm functions. They share common features but are not identical.

According to pka's view, \(\displaystyle log(a) \equiv log_e(a), \text { where } e \approx 2.718282.\)

because everybody naturally thinks in terms of integral calculus, including kids being introduced to logarithms for the first time.

According to the more old fashioned convention,

\(\displaystyle log(a) \equiv log_{10}(a) \text { and } ln(a) \equiv log_e(a)\)

I find that ancient rule better for a number of reasons such as that I personally find it easier to compute powers of 10. (Hey, I am not a mathematician so I can admit to my reliance on counting by my fingers.)

In either case, for any other base, you must specify the base of the logarithm. Otherwise, no one has a clue which logarithm function you are talking about. Remember: there are an infinite number of different ones.

\(\displaystyle u = 2^{(n-1)} \text { and } d = 3^u \implies\)

\(\displaystyle log_2(u) = log_2(2^{(n - 1)}) = n - 1 \text { and } \log_3(d) = log_3(3^u) = u \implies\)

\(\displaystyle log_2\{log_3(d)\} = log_2(u) = n - 1 \implies\)

\(\displaystyle n = 1 + log_2\{log_3(d)\}.\)

NOTE: This is a virtual duplicate of my response to the same question asked in another thread.
 
@Dr. Peterson- Ah. I didn't realize it was a function. So with equality and inequality statements, you can apply the same function to both sides the same way you could apply the same addition, subtraction, multiplication, ect.. to both sides? That as long as you are altering both sides of the statement the same way, it is okay to do so?

@JeffM- I often didn't put the base down because I had seen examples of applying the log function to both sides without a specified base. In those cases, do you think the reader was supposed to assume what the base was? I will definitely specify the base from now on.
 
@Dr. Peterson- Ah. It hadn't occurred to me that I could apply whatever logarithmic function I needed to simplify the statement. I can use a logarithmic function with whatever base I want as long as I apply the same function to both sides right?

I arrived at your answer as well, Dr. Peterson. Here is my work:

D <= 3^(2^(N-1))

log3D <= log33^(2^(N-1))

log3D <= 2^(N-1) * log33

log3D <= 2^(N-1) * 1

log3D <= 2^(N-1)

log2(log3D) <= log22^(N-1)

log2(log3D) <= (N-1)*log22

log2(log3D) <= (N-1)* 1

log2(log3D) <= (N-1)

N >= log2(log3D) + 1
 
@Dr. Peterson- Ah. It hadn't occurred to me that I could apply whatever logarithmic function I needed to simplify the statement. I can use a logarithmic function with whatever base I want as long as I apply the same function to both sides right?

I arrived at your answer as well, Dr. Peterson. Here is my work:

D <= 3^(2^(N-1))

log3D <= log33^(2^(N-1))

log3D <= 2^(N-1) * log33

log3D <= 2^(N-1) * 1

log3D <= 2^(N-1)

log2(log3D) <= log22^(N-1)

log2(log3D) <= (N-1)*log22

log2(log3D) <= (N-1)* 1

log2(log3D) <= (N-1)

N >= log2(log3D) + 1

Exactly.

You can skip a couple of those steps if you want; for instance, log33^(2^(N-1)) can immediately become 2^(N-1) because of the fact that log33x = x. (That is, the log is the inverse of the corresponding exponential function, so that the log is just the exponent.)

Also, have you seen that, via the change of base formula, the answer you got before is the same as this answer? For example, here we have log3D; that's the same as logD/log3, regardless of what base you take that to be.

And as you say, the important thing is that the log you use on both sides must be the same function (that is, must have the same base).
 
I will keep those rules in mind. Thank you @Dr. Peterson. I feel like I have a much better grasp of logs now.
 
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