What is the assumed base when applying logs to both sides?

[vaderboi]

say:

x < 5^y

and I wanted to apply log to both sides:

log x < log 5^y

What would the base be for each logarithm? Can we assume it is 10? Or is there nothing we can assume about it?

 

[Deleted member 4993]

say:

x < 5^y

and I wanted to apply log to both sides:

log x < log 5^y

What would the base be for each logarithm? Can we assume it is 10? Or is there nothing we can assume about it?

The base can be any real number as long as you use same base on both sides.

When the bases are not "explicitly" defined, the assumption is that the bases are 10.

 

[Dr.Peterson]

say:

x < 5^y

and I wanted to apply log to both sides:

log x < log 5^y

What would the base be for each logarithm? Can we assume it is 10? Or is there nothing we can assume about it?

You can choose what base to use; base 5 is convenient for some purposes, 10 or e for others.

 

[JeffM]

say:

x < 5^y

and I wanted to apply log to both sides:

log x < log 5^y

What would the base be for each logarithm? Can we assume it is 10? Or is there nothing we can assume about it?

This is a very good question.

Many decades ago, when logarithms were still heavily used for accurate computation, it was a very general convention that when a base was not specified, what was intended was a base of 10.

$\displaystyle \text {By convention, } log(x) \equiv log_{10}(x).$.

When using calculus, what is frequently useful are logarithms to the base e, what are called natural logarithms. A special notation was universal:

$\displaystyle ln(x) = log_e(x).$

Because logarithms to the base 10 are much less frequently of practical use today, some people now have adopted a different convention, namely

$\displaystyle \text {By convention, } log(x) \equiv log_e(x).$

I dislike that modern and far from universal convention for several reasons, one being that typing "log" requires 50% more work and space than typing "ln." Another reason is that it misses the importance of potential longevity of written communication.

I recommend sticking with the older convention, which is less work and avoids ambiguity.

None of this, however, is relevant to your immediate problem. You have an exponential inequation.

You want to get rid of the base of the exponent in an equation. The way to do so is to use a logarithm to that specific base. If that specific base is anything but e or 10, you must explicitly show it as the base for the logarithm. You always apply logarithms of the same base to both sides of the equation.

$\displaystyle a > 0,\ a \ne 1, \text { and } b = a^x \implies$

$\displaystyle log_a(b) = log_a(a^x) = x * log_a(a) = x * 1 = x.$

Ignoring all the technical qualifications

$\displaystyle b = a^x \iff x = log_a(b).$

How do you apply this to getting a decimal approximation for x?

You use the change of base formula.

Now try to do your problem, show us your work, and we can help from there.

EDIT: By the way, the almost exclusive practical function of logarithms to the base 10 today is to convert the solutions of exponential equations (and inequations) to decimal approximations.

EDIT 2: The internal arithmetic of computers and calculators is partially based on logarithms, most commonly to the base 16 today, so hardware designers have practical use for the theory of logarithms. But that is very much a modern practical use for only a handful of highly trained specialists.

 

[vaderboi]

Thanks for all of the responses everyone!

So would it go like this?:

log x < log 5^y

log x < y * log 5

log x / log 5 < y

x / 5 < y OR x < 5y

Does dividing a log by a log cancel both of the logs? If so, what is the logic behind that?

 

[Dr.Peterson]

Thanks for all of the responses everyone!

So would it go like this?:

log x < log 5^y

log x < y * log 5

log x / log 5 < y

x / 5 < y OR x < 5y

Does dividing a log by a log cancel both of the logs? If so, what is the logic behind that?

Have you learned the change of base formula?

log_a(x) = log_b(x)/log_b(a)
​

This is what a quotient of logs is -- it doesn't "cancel" logs, but makes a new log. So what you get on your left side is not x/5 but log_5(x).

If you were asked for the exact solution, that is it. And you could have found it by taking the base 5 log:

log₅ x < log₅ 5^y
log₅ x < y
y > log₅ x
​

On the other hand, if you were asked for a decimal approximation, then you can stop your work at

y > log x / log 5
​

and evaluate this using either base 10 or base e (since your calculator will do those directly).

Or, some calculators can evaluate log₅ x directly, so you could use the base 5 approach.

This is what I meant in saying it is your choice, whichever base is convenient for your goal. You aren't reading a log that someone else wrote, and trying to figure out what base they had in mind; you are deciding for yourself which way to do it.

If you decide early which base to use, you should write the base as I did above, so you know what you are doing. If you hold off on that decision, you can just keep writing "log" meaning "whatever base I end up deciding on", and then decide what to use at the end.

But don't do what you did here and solve for x at the end; it was solved for x in the first place. The goal was to solve for y!

EDIT: In an inequality, you can't entirely ignore the base; in your work, the base must be greater than 1, or the direction of the inequality would change (because the log would be a decreasing, rather than increasing, function). But we normally use bases like 10, e, or 5, which don't cause trouble.

 

[Deleted member 4993]

Thanks for all of the responses everyone!

So would it go like this?:

log x < log 5^y

log x < y * log 5

log x / log 5 < y

x / 5 < y OR x < 5y

Does dividing a log by a log cancel both of the logs? If so, what is the logic behind that?

NO .... NO .... Double NO!!!

Just like while dividing 26 by 6

You CANNOT cancel off 6's

and declare the answer to be 2.

 

[vaderboi]

Thanks @Subhotosh Khan. So log cannot be simplified by dividing by another log.

So y > log x / log 5 is as simple as it can get.

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

I bring up this example because it was another problem that I was struggling with in terms of simplifying logs. I can ask this in another thread if need be.

 

[pka]

say: x < 5^y
and I wanted to apply log to both sides: WHY?
What would the base be for each logarithm?

What is the point of this?
Do you understand $\displaystyle (\forall y)\left[5^y>0\right]~?$
So $\displaystyle \{(x,y): x\le 0\}$ is a subset of the solution for $\displaystyle x<5^y$
What other pairs satisfy relation? SEE HERE
At the wolframalpha site you can see $\displaystyle x>0~\&~y>\dfrac{\log(x)}{\log(5)}$

That last bit brings back to what is understood by $\displaystyle \log(x)$.
Mathematics education is driven by two organisations: the MAA(Mathematics Association of America) & the NCTM(National Council of Teachers of Mathematics) Leonard Gillman was the president of the MAA and published a seminal calculus textbook in 1973.
In that text he gave the definition of the logarithm function as: If $\displaystyle x>0$ then $\displaystyle \displaystyle{\log (x) = \int_1^x {\frac{1}{t}} dt}$
Because this approach was so natural it is easy show that $\displaystyle \log(xy)=\log(x)+log(y)$ and to define $\displaystyle \log (e)=1$.
So it has become standard for most computer algebra systems(CAS) and/or calculators to use $\displaystyle \log(x)$ in this natural way.
Now this is a correction: a logarithm base must be positive. So we now have $\displaystyle \log_b(A)=\dfrac{\log(A)}{\log(b)}.$

 

[Otis]

What is the point of this?
Do you understand [?] …

I was thinking of asking you the same questions!

 

[vaderboi]

@pka- I do not understand pretty much any of what you said. Hahaha. I am very new to logarithms.

 

[JeffM]

Thanks @Subhotosh Khan. So log cannot be simplified by dividing by another log.

So y > log x / log 5 is as simple as it can get.

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

I bring up this example because it was another problem that I was struggling with in terms of simplifying logs. I can ask this in another thread if need be.

Did you even read Dr. Peterson's or my posts?

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.

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

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

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

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

Now, as Dr. Peterson mentioned, inequations involving logarithms with bases less than 1 introduce a wrinkle, but it does not apply to this problem.