Is "ln" (natural log) and "log" the same thing if used in this answer? Find $x$ for $4^{x-4} = 7$.
Answer I got, using log, was ${\log(7)\over 2\log(2)} + 4$
but the actual answer was ${\ln(7)\over2\ln(2)} + 4$
I plugged both in my calculator and turns out both are the equivalent value.
Anyways, is using either one of ln or log appropriate for this question? Obviously ln is when log has the base e, and log is when it has the base 10. 
Final question: How do I know when to use which? that is which of ln or log is used when solving a question??
For example, if a question asks to find $x$ for $e^x = 100$, I will use $\ln$ since $\ln(e)$ cancels out.
If a question asks to find $2^x = 64$, i will use log since "$e$" isn't present in the question.
So is using either $\log$ or $\ln$ the same?
 A: Culturally


*

*Computer science / programming people tend to use log base $2$

*Mathematicians tend to use log base $e$

*Engineers / physicist / chemists etc. tend to use log base $10$
Writers really should make it explicit the first time they use "$\log$", but they don't always. As others have pointed out, the only difference is a constant factor, and in your case the factors in the numerator and denominator cancelled each other out. So the answer to your question is "If it's in a math context you'll probably see $\ln()$ used." 
Heck, even if you were asked to solve $10^{x - 3} = 6$ you'd still see $\ln()$ used, even though it looks like $\log_{10}$ might seem more "natural" for that particular problem. It's just what math people tend to do.
A: There's an interesting unstated question here: what counts as an answer?
You can clearly argue that using either $\ln$ or $\log_{10}$ should be acceptable. But in that case 
$$
x = \log_4(7) + 4
$$
should be just as correct. As @BrianMoehring says in his answer, you can use any logarithm you want.
As for

If a question asks to find $2^x=64$, i will use log since "e" isn't
  present in the question.

I would just say $x=6$. That's really using $\log_2$, by inspection.
A: You can use any logarithm you want.
As a result of the base change formula
$$\log_2(7) = \frac{\log(7)}{\log(2)} = \frac{\ln(7)}{\ln(2)} = \frac{\log_b(7)}{\log_b(2)}$$
so as long as both logs have the same base, their ratio will be the same, regardless of the chosen base (as long as $b > 0, b\neq 1$).
A: In this case the other two answers are technically correct of course, in that the base doesn't really matter. But I want to point out that when you see $\log(x)$, it can either mean base $10$, base $2$ or base $e$, with the latter two (especially base $e$) being much more common as you move up the math ladder. The notation $\ln(x)$ is still used for base $e$, but whenever you see $\log(x)$ you should always assume it is also base $e$ unless context implies otherwise (if it's supposed to mean base $2$, it should be clear from context).
Part of the reason is exactly because of the reason mentioned by the two other answers: for any $a,b$ we have
$$\log_a(b)=\frac{\log(b)}{\log(a)},$$
So we can express logarithms of any base using the natural logarithm anyway and there's no need to designate a special symbol for it. And indeed you will see that base $e$ is much more useful than base $10$ most of the time.
A: Either is fine.  You can write logarithms in terms of any base that you like with the change of base formula
$$\log_ba=\frac{\log_ca}{\log_cb}$$
One thing learned from secondary school is that exponential equations can be solved be rewriting the relationship in logarithmic form.  As such, your equation can be rewritten as follows:
\begin{align}
4^{x-4}&=7\\
x-4&=\log_47\\
x&=4+\log_47
\end{align}
And since $\log_47$ can be rewritten as $\frac{\log7}{\log4}$ or $\frac{\ln7}{\ln4}$ or $\frac{\log_{999876}7}{\log_{999876}4}$ it does not matter which base of logarithm you use.
A: Use the change of base formula if you suspect that the answer is irrational, otherwise take the logarithm to both sides of the equation of a base that seems reasonable. 
There's nothing magical about the change of base formula.
$$\begin{align}
\log_c b &= \log_c b\\
\log_c a^{\log_a b} &= \log_c b\\
\log_a b \cdot\log_c a &= \log_c b\\
\log_a b &= \frac{\log_c b}{\log_c a}\\
\end{align}$$
Even if the solution is integral or rational, using  the change of base formula will get you to an answer, for example:
$$\begin{align}
2^x &= 64\\
\log_{10}2^x &= \log_{10} 64\\
x\log_{10}2 &= \log_{10} 64\\
x &= \frac{\log_{10} 64}{\log_{10}2}\\
x &= \log_{2} 64\\
x &= \log_{2} 2^6\\
x &= 6\log_{2} 2\\
x &= 6\\
\end{align}$$
although, it does insert a lot of extra [unnecessary] steps.
